Projectile question with no time or velocity or range

In summary, the problem involves a fly ball that clears a 10.0 meter fence at a 45 degree angle and strikes the ground 8.0 meters beyond the fence. The equation dy=vit+1/2at2 can be used to solve for the speed of the ball when it left the bat, but there is not enough information given. The solution lies in finding the initial speed of the ball at the top of the wall, which can be determined by writing equations of motion for the x and y components and using the known distances and angles. Once the initial speed is found, the impact speed at the ground can be calculated.
  • #1
Ariesspear
8
0

Homework Statement



A fly ball just clears a 10.0 meter fence on the way down at a 45 degree angle, and strikes the ground 8.0 meters beyond the fence. Calculate the speed of the ball when it left his bat (at ground level).

since it lands at ground level dy=0
a=9.81m/s2 since it is the only acceleration acting on a projectile.

Homework Equations



dy=vit+1/2at2

v=dx/t

The Attempt at a Solution


Ok so I've been all over the place with this question and i have no idea how to solve it. There is no mass so i cannot use any type of energy formula, no time which means i will at some point have to find it, no velocities to start off with, and no overall range or dx distance.
If i had the dx distance then i would be able to solve it since i would have 3 of 4 necessary info for the dy=vit+1/2at2 equation, but without it I am lost.
 
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  • #2
Ariesspear said:

Homework Statement



A fly ball just clears a 10.0 meter fence on the way down at a 45 degree angle, and strikes the ground 8.0 meters beyond the fence. Calculate the speed of the ball when it left his bat (at ground level).

since it lands at ground level dy=0
a=9.81m/s2 since it is the only acceleration acting on a projectile.

Homework Equations



dy=vit+1/2at2

v=dx/t

The Attempt at a Solution


Ok so I've been all over the place with this question and i have no idea how to solve it. There is no mass so i cannot use any type of energy formula, no time which means i will at some point have to find it, no velocities to start off with, and no overall range or dx distance.
If i had the dx distance then i would be able to solve it since i would have 3 of 4 necessary info for the dy=vit+1/2at2 equation, but without it I am lost.
be careful to use enough adjectives to indicate what portion of the trajectory your equation is about.
For example, you DO know some distance, 10m ... and another one, 8m ... (one of them will be used as a negative).
Draw this, so you know which is which.
It might be better to use dy = vy,f t - ½ a t2 ... this t is only after clearing the fence.
 
  • #3
lightgrav said:
be careful to use enough adjectives to indicate what portion of the trajectory your equation is about.
For example, you DO know some distance, 10m ... and another one, 8m ... (one of them will be used as a negative).
Draw this, so you know which is which.
It might be better to use dy = vy,f t - ½ a t2 ... this t is only after clearing the fence.
I set up a triangle which has the 10.0 meter fence as the y-axis and the 8.0 meter distance as the x-axis and the balls path is the hypotenuse in between. I do have those distances but without any form of time i still can't calculate anything from that triangle i don't think. And the Vy,f is the final velocity correct? which means that it would simply be zero, leaving me with 0 = -4.9t2
 
  • #4
The ball's path won't be a straight line: It follows a parabolic trajectory due to the influence of gravity.

You've recognized the problem within the problem, namely the trajectory from the top of the wall to the ground:

Fig1.gif


If you assume some speed for the ball at the top of the wall, say vw, and the projectile is moving at 45° below the horizontal from the top of the wall, and you know that it strikes the ground at 8 m from the wall, can you determine the speed vw?

You should be able to write equations of motion for the x and y directions, and write the initial velocities in the x and y directions in terms of vw (sin(45°) and cos(45°) have simple values). There's enough information to find vw.

Once you know the speed vw at the top of the wall you should be able to find the impact speed at the ground.
 
  • #5
gneill said:
The ball's path won't be a straight line: It follows a parabolic trajectory due to the influence of gravity.

You've recognized the problem within the problem, namely the trajectory from the top of the wall to the ground:

View attachment 77962

If you assume some speed for the ball at the top of the wall, say vw, and the projectile is moving at 45° below the horizontal from the top of the wall, and you know that it strikes the ground at 8 m from the wall, can you determine the speed vw?

You should be able to write equations of motion for the x and y directions, and write the initial velocities in the x and y directions in terms of vw (sin(45°) and cos(45°) have simple values). There's enough information to find vw.

Once you know the speed vw at the top of the wall you should be able to find the impact speed at the ground.
Im not sure if i understand. I know that you can find the components (such as Vx and Vy) for an angled velocity by using the equations Vx=Vcosθ and Vy=Vsinθ. I am doing this for an advanced physics assignment at school and I am the only one doing it and I've been running over all of my notes and i still can't find anything to help my case.
 
  • #6
Ariesspear said:
Im not sure if i understand. I know that you can find the components (such as Vx and Vy) for an angled velocity by using the equations Vx=Vcosθ and Vy=Vsinθ. I am doing this for an advanced physics assignment at school and I am the only one doing it and I've been running over all of my notes and i still can't find anything to help my case.
Write the kinematic equations for the x and y components of the motion from the top of the wall. Assuming an initial speed vw, what are the x and y positions of the object with respect to time?
 
  • #7
gneill said:
Write the kinematic equations for the x and y components of the motion from the top of the wall. Assuming an initial speed vw, what are the x and y positions of the object with respect to time?

Ive used the Vx=Vcosθ and Vy=Vsinθ ( where θ = 45) yet all i get from both of those is 0.7071V. You mention in respect to time and i assume that i am misinterpreting what you mean to convey, as there is no time as of right now in my equations and problem.
 
  • #8
Write the kinematic equations of the motion! y(t) = ..., x(t) = ...

Yes, you are introducing a time variable t, but you will soon get rid of it by combining the equations. So, just start by writing the equations of motion. You want to solve for the initial velocity at the top of the wall that results in the projectile striking the ground 8 m from the base of the wall. You have enough information given: the launch angle, initial height, and horizontal distance at the point of impact.

By the way, you should memorize the numerical value for sin(45°) and cos(45°) so you don't have to drag the trig functions around through all your calculations. Other convenient sin and cos values to know are those for 30° ;)
 
  • #9
gneill said:
Write the kinematic equations of the motion! y(t) = ..., x(t) = ...

Yes, you are introducing a time variable t, but you will soon get rid of it by combining the equations. So, just start by writing the equations of motion. You want to solve for the initial velocity at the top of the wall that results in the projectile striking the ground 8 m from the base of the wall. You have enough information given: the launch angle, initial height, and horizontal distance at the point of impact.

By the way, you should memorize the numerical value for sin(45°) and cos(45°) so you don't have to drag the trig functions around through all your calculations. Other convenient sin and cos values to know are those for 30° ;)


God i feel like such an idiot, I've got you helping me and I am still confused. Ok so these are the equations I've got,
d = vo • t + 0.5 • a • t2
vf = vo + a • t
vf2 = vo 2 + 2 • a • d
d = (vo + vf)/ 2 • t
and the way you put it in your post ("y(t) = ..., x(t) = ...") confuses me as i don't understand that format of the equation; which truth be told only makes me feel worse about myself.
I kind of understand what you mean by combining the two equations to cancel out the time variable, but again i can't move forward until i understand what equation to use; and i promise i am decently smart i just lack abstract thinking, i need to do or see something once before i can understand it or repeat it.
 
  • #10
y(t) means y as a function of t, x(t) means x as a function of t.

From the assumed initial speed and given launch angle you can write the initial x and y velocity components. Then you can write the equations for the x position and y position with respect to time using those initial velocities. Those are the basic projectile motion equations.

So, write the equation for the y position given the initial y-position, initial y-velocity and the known acceleration due to gravity. Also write the x-position given the initial x-position and x-velocity (assume that the wall is at x position 0 m).
 
  • #11
math notation writes "f, as a function of x" like f(x).
physics folks sometimes use math notation, but more often use subscripts, like your vf ... final is a time, after all.
Almost always, we use x as horizontal; for vertical we either use y or z.

Notice that your "d" equation has an acceleration ... if the acceleration is NOT zero, which direction is it? (x or y) ... call it that (dx or dy)
If the acceleration IS zero, which direction is it? (x or y) ... call it that (dx , or dy)
Now, solve the simpler equation for "t" ... and plug it (that symbolic formula) into the nastier equation.
 
  • #12
lightgrav said:
math notation writes "f, as a function of x" like f(x).
physics folks sometimes use math notation, but more often use subscripts, like your vf ... final is a time, after all.
Almost always, we use x as horizontal; for vertical we either use y or z.

Notice that your "d" equation has an acceleration ... if the acceleration is NOT zero, which direction is it? (x or y) ... call it that (dx or dy)
If the acceleration IS zero, which direction is it? (x or y) ... call it that (dx , or dy)
Now, solve the simpler equation for "t" ... and plug it (that symbolic formula) into the nastier equation.

ok that makes some sense to me, so the y(t) = ..., x(t) = ... gneill was talking about refers to dx =...and dy =... equations? such as dy = vi • t + 0.5 • a • t2
And i am assuming that the "nastier equation" is referring to the formula that i create to cancel out a variable that i don't have yet?

The d equation with acceleration is the dy (vertical) as it is gravity that acts upon it. and the one without acceleration (=0) is horizontal.
 
  • #13
yeah, go for it! by coincidence, in this problem the vx = -vy , only as the ball crosses the fence
... for the vertical one vfence,y =/= vground,y
 
  • #14
Ok well ill give it a shot tomorrow and post back on here if I've gotten it or not.
 
  • #15
Ok so i understand what you mean about vx=-vy, so with this being said i should be able to take dx/t = the rearranged formula for dy=vit+1/2at2, my only issue is that it doesn't work out. I am assuming that I am rearranging something wrong or my logic is flawed.
 
Last edited:

1. What is a projectile?

A projectile is any object that is launched or thrown through the air and is subject to the forces of gravity and air resistance. Examples of projectiles include a baseball thrown by a pitcher or a cannonball fired from a cannon.

2. Can a projectile have no initial velocity or time?

Yes, a projectile can have no initial velocity or time. This means that the projectile is not launched or thrown, but simply dropped from a certain height. In this case, the initial velocity is 0 m/s and the initial time is 0 seconds.

3. How do you calculate the range of a projectile with no initial velocity or time?

The range of a projectile with no initial velocity or time can be calculated using the equation R = (1/2) * g * t^2, where R is the range, g is the acceleration due to gravity (9.8 m/s^2), and t is the time the projectile is in the air. This equation assumes that the projectile is launched from a horizontal surface and lands on the same horizontal surface.

4. What is the role of air resistance in a projectile with no initial velocity or time?

Air resistance, also known as drag, is a force that acts in the opposite direction of the motion of a projectile. In a projectile with no initial velocity or time, air resistance affects the time it takes for the projectile to fall to the ground and therefore, the range of the projectile.

5. Can a projectile with no initial velocity or time be affected by external forces?

Yes, a projectile with no initial velocity or time can still be affected by external forces such as wind or a change in gravity. These external forces can alter the trajectory and range of the projectile.

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