What Height Was the Ball Thrown From in This Projectile Motion Problem?

[Star Forger]

Homework Statement

A ball thrown horizontally at 24 m/s travels a horizontal distance of 49 m before hitting the ground. From what height was the ball thrown?

Homework Equations

The Attempt at a Solution

I'm not sure how to start this. Could someone please help?

 

[cepheid]

You know how long it was in the air, because you have both the horizontal speed (which is constant) and the horizontal distance travelled. From this time spent in the air, you can deduce what height it must have fallen from (because in the vertical direction, the ball is just in free fall).

 

[Star Forger]

Ok, I got the time. Now is there a specific equation I should use? I'm guessing it's just a variation of one of the Kinematic Equations?

 

[cepheid]

Free fall means under the influence of gravity only. Hence acceleration is constant. This is exactly the condition in which the kinematics equations apply. To choose the correct one, consider your givens (in the vertical direction). You have been given the time, acceleration, and initial velocity, and you need to solve for the distance.

 

[Star Forger]

Thank you so much! I've just got one more question: Since we are considering the vertical direction, acceleration due to gravity is still -9.8, not +9.8, correct?

 

[cepheid]

It's up to you. You can pick either sign convention: "downward is negative," or, "downward is positive," so long as you stick to it consistently throughout the problem.

 

[Star Forger]

Ok. I thought I did this correctly, however when I entered my answer of 28.6 it said that it was incorrect. I used d = vi*t + .5*a*t^2, where vi = 24, t = 2.04, and a = -9.8. Do you know where I went wrong in my calculations?

 

[cepheid]

Ok. I thought I did this correctly, however when I entered my answer of 28.6 it said that it was incorrect. I used d = vi*t + .5*a*t^2, where vi = 24, t = 2.04, and a = -9.8. Do you know where I went wrong in my calculations?

It doesn't seem like you're quite thinking things through here. Like with any projectile motion problem, we can consider the horizontal and vertical motions independently. I'll refer to the horizontal position coordinate as 'x' and the vertical one as 'y' to distinguish the two distances. This is really standard notation. For the x-direction, there is no acceleration, since the only force that acts is gravity, and it acts entirely vertically (in the y-direction). Hence, horizontal speed is constant, and the equation for distance vs. time is:

x = v_xt

where v_x = 24 m/s (given).

You used this to solve for t, which was the total travel time. (If it had been in the air longer, it would have gone farther horizontally, and x would be correspondingly larger).

In the y-direction, there IS acceleration, due to gravity. Hence, we can write a_y = -g, where g = +9.81 m/s², and I have chosen downward to be the negative y-direction. So, the formula for distance vs. time in this direction is:

y = v_iyt - (1/2)gt²

The initial vertical velocity (v_iy) is NOT 24 m/s (this is the horizontal velocity). However, you know what v_iy is from the situation. Hint: the thrower throws the ball entirely horizontally. This means that at the instant the ball is released, its only velocity is horizontal.