Projectile Motion - Seems easy.

In summary: Thank you for your help. :)In summary, the conversation discusses the concept of projectile motion and solving for various components of a projectile's motion, such as time, range, velocity, and angle. The conversation also includes a mistake in solving for time, which is later corrected.
  • #1
Kildars
95
0
Projectile Motion -- Seems easy.

A projectile is shot from the edge of a cliff 245 m above ground level with an initial speed of vo=135 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-44.

3-44alt.gif


Figure 3-44

(a) Determine the time taken by the projectile to hit point P at ground level.
s

(b) Determine the range X of the projectile as measured from the base of the cliff.
km

(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.)
m/s (horizontal)
m/s (vertical)

(d) What is the magnitude of the velocity?
m/s

(e) What is the angle made by the velocity vector with the horizontal?
° below the horizontal

I tried doing \Delta X = Vox*t for A but I know that won't work because 245 is \Delta Y I can't solve for time without either \Delta X or Voy which is also what I don't have.. any hints?
 
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  • #2
Sorry I don't know how to enter equations here, so I'll do the best I can.

Lets start with;
y=((v0) sin(θ))t - 1/2gt^2
defines the y location of the projectile. Set this to ground level and we have;
-245 = (135 sin(37))t - 1/2(9.8)t^2
0 = 245 + 81.245*t - 4.9*t^2

I don't have a calculator handy, but solve that for t and you know the time it took for the projectile to hit the ground.

x=((v0) cos(θ))t give the x location if you enter the time value from above.

At impact the horizontal component is the same as when it left (assuming no air resistance);
v(x) = (v0)cos(θ)

The y component is;
v(y) = (v0)sin(θ) - gt

velocity at impact is;
v = sqr( (v(x))^2 + (v(y))^2 )

lastly, the angle at impact is;
θ = arctan( v_y/v_x) )

I'm a student also, so please check my work. Also, how can formula's be better input so they look nicer?

Bernie
 
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  • #3
-245 = 81.245*t - 4.9t^2

I got .784.. I got that by

-245/8.245 then dividing the quotient of that by -4.9.. then square rooting that.

Which is incorrect.

I see this is your first post, thank you for your help.. if I did something wrong let me know.
 
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  • #4
To make formulas look better you type "[..tex..]" formula stuff "[../tex..]" without the quotes or the periods.
 
  • #5
Kildars said:
-245 = 81.245*t - 4.9t^2

I got .784.. I got that by

-245/8.245 then dividing the quotient of that by -4.9.. then square rooting that.

Which is incorrect.

I see this is your first post, thank you for your help.. if I did something wrong let me know.
Bernie's method and solution is correct. You messed up on solving for t. This is a quadratic equation. Solve for t using the quadratic formula.
 
  • #6
PhanthomJay said:
Bernie's method and solution is correct. You messed up on solving for t. This is a quadratic equation. Solve for t using the quadratic formula.

Ah lame! Alright let me try.
 
  • #7
Okay I got 19.187 for A and it worked. I then did the formulas to get B and C

B was 2.068km
C was 107.815..

I'm stuck on D though

for v(y) = (v0)sin(θ) - gt

I don't know what to do with that

I tried doing 135sin(37) - 9.81(19.187)

and that does not work.
 
  • #8
Hmm... in letter C, you are asked to get the x & y components... so once you got the components... get the magnitude by V=sqrt(Vx^2+Vy^2)

to get the angle, just get the inverse tangent of the quotient of the y-component and x-component

because tanθ=Vy/Vx
 
  • #9
Yeah for C2, the second C.. I can't get the component with the formula he gave me.. it doesn't add up right.
 
  • #10
C2? The x component, you mean? It stays the same--the vertical component (v_x) is constant at all point, so v_x = v_fx = v_ix. Just plug that in the formulas Bernie gave you and you should get the right answer. And I believe you incorrectly calculated (a) if you got 19.817--you should get 10.63 seconds.
 
  • #11
johnnyp said:
C2? The x component, you mean? It stays the same--the vertical component (v_x) is constant at all point, so v_x = v_fx = v_ix. Just plug that in the formulas Bernie gave you and you should get the right answer. And I believe you incorrectly calculated (a) if you got 19.817--you should get 10.63 seconds.

No, my webassign automatically checks the answer and it said it's correct at 19.87. X is horizontal, Y is vertical.
 
  • #12
Okay I got c2 using bernie's formula. Thanks. :)

Trying the rest.
 
  • #13
Very much sorry, Kildars. I neglected the (1/2) in the equation. :blushes: Good luck with the problem!
 
  • #14
Got them all, thanks for the help guys.
 
  • #15
johnnyp said:
Very much sorry, Kildars. I neglected the (1/2) in the equation. :blushes: Good luck with the problem!

No problem!
 

Related to Projectile Motion - Seems easy.

1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It is a combination of horizontal and vertical motion, and it follows a curved path known as a parabola.

2. How is projectile motion different from regular motion?

Regular motion follows a straight path, while projectile motion follows a curved path due to the influence of gravity. In regular motion, the speed and direction of the object remain constant, while in projectile motion, the speed changes continuously due to the acceleration of gravity.

3. What are the factors that affect projectile motion?

The factors that affect projectile motion are the initial velocity, the angle of launch, the air resistance, and the acceleration due to gravity. These factors determine the shape and distance of the projectile's trajectory.

4. What is the formula for calculating the maximum height of a projectile?

The formula for calculating the maximum height of a projectile is h = (v02 sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

5. How can projectile motion be applied in real life?

Projectile motion has many real-life applications, such as in sports like basketball and football, where players need to calculate the trajectory of the ball to make successful shots. It is also used in military and engineering to calculate the trajectory of missiles and rockets. Additionally, it is applied in physics experiments to study the effects of gravity on objects in motion.

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