Calculating a Projectile's Max Height, Time & Range

In summary, a projectile is launched at an angle of 45 degrees with a velocity of 200m/s. To calculate the maximum height reached, use the equation v^2=u^2+2as. To calculate the time of flight, use the equation s=ut+1/2at^2. To calculate the range, use the equation s=vt or v^2/g. It is important to use notation such as Vx and Vy to keep things clear, and to solve simultaneous equations for x(t) and y(t) in order to determine the time of impact at the far end. It is also recommended to avoid giving out canned formulas and instead provide tutorial hints for the OP to better understand how to approach similar problems
  • #1
Awsom Guy
77
0

Homework Statement


A projectile os launched with velocity of 200m/s at an angle of 45deg to the horizontal. Calculate:
a) the maximum height reached
b) the time of flight
c) the range


Homework Equations


v^2=u^2+2as
s=ut+1/2 at^2


The Attempt at a Solution


I know the first 2 but c) I am very confused.
s=(u cos theta)
t=(200 cos 45)

Please give me a hand with c)
Thanks
 
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  • #2
Awsom Guy said:

Homework Statement


A projectile os launched with velocity of 200m/s at an angle of 45deg to the horizontal. Calculate:
a) the maximum height reached
b) the time of flight
c) the range


Homework Equations


v^2=u^2+2as
s=ut+1/2 at^2


The Attempt at a Solution


I know the first 2 but c) I am very confused.
s=(u cos theta)
t=(200 cos 45)

Please give me a hand with c)
Thanks

Your notation is confusing -- it would be easier to call the components of V by something intuitive like Vx and Vy.

In this type of problem, the usual approach is to use the fact that Vx is constant (no gravity that way), and you only have an acceleration component in the Vy direction. Calculate the time of flight based on the fact that your projectile falls back down to y=0 at some time t, and solve for t. Then since Vx is constant, you can ...
 
Last edited:
  • #3
That doesn't make sense sorry.
 
  • #4
Awsom Guy said:
That doesn't make sense sorry.

Which part doesn't make sense? Start by writing the kinematic equations of motion using Vx and Vy notation to keep things clear.

x(t) =
y(t) =

Vx(t) =
Vy(t) =

Then write the two simultaneous equations for the x and y motion of the projectile:

x(t) = (has to do with the Vx(t) which is constant, and time)
y(t) = (has to do with the Vy(t) which depends on Vy(0) and gravity g and time)

Solve the 2 simultaneous equations for when the projectile hits the ground at the far end (hint -- what does y= when it hits the ground). That gives you the time of impact. The rest is gravy...
 
  • #5
A projectile os launched with velocity of 200m/s at an angle of 45deg to the horizontal. Calculate:
c) the range

To figure out range you can just use: s = vt or << answer deleted by berkeman >>
 
  • #6
Sirsh said:
A projectile os launched with velocity of 200m/s at an angle of 45deg to the horizontal. Calculate:
c) the range

To figure out range you can just use: s = vt or << answer deleted by berkeman >>

It's okay to help, but it's not okay to give out an answer. Please re-check the Rules link at the top of the page, particularly the part about Homework Help.

This thread may also be instructive:

https://www.physicsforums.com/showthread.php?t=373889

.
 
  • #7
s=vt or what?
 
  • #8
Why was a formula deleted? How's that giving an answer
 
  • #9
Please give me the formula Sirsh.
 
  • #10
I won't give you it directly. but if you have a look on wikipedia under trajectory or trajectory of a projectile. it should be there
 
  • #11
But there are so many equations which one is right :(.
 
  • #12
oh don't worry found it :D
Thanks
 
  • #13
Is this eqaution good for finding the answer to c),
s=v^2/g.
 
  • #14
That could work. don't forget the rest of the equation just incase you get a problem, as those equations in comparison to the 2D motion ones are always off by around a tenth place. just stick with s=vt, if you have found the time then you should be fine with that equation.
 
  • #15
Thanks, that makes sense.
 
  • #16
No problem mate.
 
  • #17
Sirsh said:
Why was a formula deleted? How's that giving an answer

You did the right thing later in this thread. Instead of handing out a formula, hint at what it might be and mention a resource where they might find it. Making the OP do more of the work helps them learn how to solve problems better on their own in the future.

Even better yet, don't hint at a canned formula, give tutorial hints at how to solve the problem from the basics. That will generally help the OP's understanding of how to approach these general types of problems better in the future on their own (like on tests).
 
  • #18
Oh okay, I'll do that instead then :) also, what does OP mean?
 
  • #19
OP is original poster, the person who started the thread with the original question. Thanks.
 
  • #20
oh okay. thank you.
 
  • #21
Thankyou guys.
 

1. How do you calculate the maximum height of a projectile?

The maximum height of a projectile can be calculated using the formula h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

2. What is the formula for calculating the time of flight of a projectile?

The time of flight of a projectile can be calculated using the formula t = 2v sinθ/g, where t is the time of flight, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

3. How do you determine the range of a projectile?

The range of a projectile can be determined using the formula R = v2sin2θ/g, where R is the range, v is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

4. Can the maximum height, time of flight, and range of a projectile be calculated if the initial velocity and angle of launch are not given?

No, all three values cannot be calculated without knowing the initial velocity and angle of launch of the projectile. At least one of these variables is necessary to calculate the maximum height, time of flight, and range.

5. How does air resistance affect the calculations for a projectile's maximum height, time of flight, and range?

Air resistance can affect the calculations for a projectile's maximum height, time of flight, and range by decreasing the initial velocity and altering the trajectory of the projectile. This can result in a lower maximum height, shorter time of flight, and shorter range compared to calculations without air resistance.

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