Projectile Motion at an angle

  • #1
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Homework Statement:
A ball is thrown from a height of 30.0m with a speed of 10.0m/s at an angle of 23.0 above the horizontal. How fast is the ball going when it hits the ground?
Relevant Equations:
Vf^2 = Vi^2 + 2a(Delta X)
Delta Y = Vit + 1/2at^2
Vx = 10cos(23) = 9.21m/s
Vy = 10sin(23) = 3.91m/s

-30 = 3.91t + 1/2(-9.8)t^2
0 = t(3.91 - 4.9t) + 30
4.9t = 33.91
t = 6.92s

Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2
Delta X = 298.38m

Vf^2 = 9.21 + 2(9.8)(298.38)
Vf^2 = 5857.458
Vf = 76.53m/s
 

Answers and Replies

  • #2
collinsmark
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Hello @bluemyner,

Welcome to PF! :welcome:

Homework Statement: A ball is thrown from a height of 30.0m with a speed of 10.0m/s at an angle of 23.0 above the horizontal. How fast is the ball going when it hits the ground?
Homework Equations: Vf^2 = Vi^2 + 2a(Delta X)
Delta Y = Vit + 1/2at^2

Vx = 10cos(23) = 9.21m/s
Vy = 10sin(23) = 3.91m/s

-30 = 3.91t + 1/2(-9.8)t^2
0 = t(3.91 - 4.9t) + 30
4.9t = 33.91
t = 6.92s
I got a different value for [itex] t [/itex].

But before we correct that, why are you solving for [itex] t [/itex] at all?

Isn't there a different approach (different equation) that you can use to solve for the final vertical velocity component that doesn't involve [itex] t [/itex]?

Delta X = 9.21(6.92) + 1/2(9.8)(6.92)^2
Delta X = 298.38m

Vf^2 = 9.21 + 2(9.8)(298.38)
Vf^2 = 5857.458
Vf = 76.53m/s

Now you lost me with all that. The ball is not accelerating in the horizontal direction, so why would you use an equation that involves acceleration for that direction?
 
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  • #3
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Thanks for the welcome! I'll be completely honest, I am very lost with this problem and I'm new to Physics forums so I saw that it said to try and answer the problem and I gave it my best shot. I've been trying the plug and chug approach to this problem and it hasn't worked out for me.
 
  • #4
collinsmark
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Thanks for the welcome! I'll be completely honest, I am very lost with this problem and I'm new to Physics forums so I saw that it said to try and answer the problem and I gave it my best shot. I've been trying the plug and chug approach to this problem and it hasn't worked out for me.
That's fine. You've come to the right place. :smile:

Before I begin, I should point out that there is more than one way to solve this problem. Given your attempt, I'm going to assume that you are supposed to use "kinematics" to solve this. (The other method is "conservation of energy." But if that's not familiar to you yet, you'll get to that method in the future.)

In order to solve this problem using kinematics, you need to find the horizontal and vertical components of the velocity separately (you may combine them at the very end).

What can you tell me about the horizontal component of velocity?

[Edit: Here's a hint. You've already answered it near the very beginning of your attempted solution. But, since there's no acceleration in the horizontal direction, what can you tell me about how the horizontal component of velocity changes with time (if it changes at all)?]
 
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  • #5
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I looked at the problem with a fresh set of eyes this morning. I resolved for time using the quadratic equation and got t = 2.91s and then solved for Final velocity in the y direction which gave me 24.56m/s. Using this and the initial velocity in the x direction and the Pythagorean theorem I found Final Velocity in the x-direction to be 26.23m/s. Does that seem right?
 
  • #6
PeroK
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I looked at the problem with a fresh set of eyes this morning. I resolved for time using the quadratic equation and got t = 2.91s and then solved for Final velocity in the y direction which gave me 24.56m/s. Using this and the initial velocity in the x direction and the Pythagorean theorem I found Final Velocity in the x-direction to be 26.23m/s. Does that seem right?
The final velocity looks right. I checked by using energy. The initial KE + initial GPE = final KE.

The angle is irrelevant.

Note that we are talking here about the final speed, which is the magnitude of the final velocity.
 
  • #7
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I looked at the problem with a fresh set of eyes this morning. I resolved for time using the quadratic equation and got t = 2.91s and then solved for Final velocity in the y direction which gave me 24.56m/s. Using this and the initial velocity in the x direction and the Pythagorean theorem I found Final Velocity in the x-direction to be 26.23m/s. Does that seem right?
All good - would however recommend that you confirm your answer using the alternate approach suggested in post #2 whereby there is no need to solve any quadratic for t. You just have to be very careful with signs of quantities used in the relevant formula.
 

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