Phyisics question?

Question

Can someone plz help me with this. Im not sure if you supposta set up 2 equations or what to do.

A ball is thrown up at an initial speed of 25 m/s at the same time a ball is dropped from rest at height of 15 m. How long does it take for the balls to be at the same height?

Answer 1

The first is for the ball thrown up.

Height = V1(t) - g/2 ( t^2) Starting height is zero for up traveling ball

For the Ball being dropped V1 is zero but Starting Height is 15 meters so the equation is.

Height = 15 - g/2 (t^2)

At the time the heights of the two balls are equal the equations will be equal or.

25t - (g/2) t^2 = 15 -(g/2) t^2

simplifying you have 25t = 15 or t = 0.6 secs

Which we can check by solving each of the equations and seeing that both equal 13.24 meters at 0.6 secs g = 9.8 meters/sec^2

Answer 2

Ok, I'm assuming that the ball going up is at constant velocity and there is no air resistance. Therefore, the solution must be attained by the following:

First, draw the picture. As a physicist, I can tell you that drawing the picture REALLY helps. You will need 2 kinematic equations: x = v*t (for the ball going up) and y = 1/2(a*t^2) for the ball falling. Now, "t" (time) is going to be the same in both equations. So rearrange both equations to equal "t". That gives you x/v on the left and (2y/a)^1/2 on the right. V and A are knowns/givens v = 25m/s and a = 9.8m/s^2. So the only things left unknown are x and y. We know that height when the 2 balls meet will be somewhere between 0-15 meters. So let x equal the height at which they meet. Therefore, x = 15 - y. We plug that value in for x and the new equality reads (15 - y)/ v = (2y/a)^1/2. Now, square both sides and algebracally move everything to one side so that the equation is equal to zero. Then, using the quadractic formula, solve for y. (look up quadratic formula, it's too tricky to type in) One of the solutions will be bogus and the other will be your solution to y. Then solve for x.

Jonesy

Answer 3

I would draw a graph with two curves on it. increasing time from t0 along the bottom, height going up the page. Assume that at t0 the ball that is being thrown upwards starts at 0 height ( or 2meters - whatever you think is reasonable - it's not in the question so you can make an assumption) assume the air resistance is negligible. plot one curve giving the height of the dropped ball against time. Plot the second curve giving height of the thrown ball against time. Note the time(s) that they are both at the same height.

Best of luck - Mike