Solving Baseball Throw: Initial Speed, Altitude & Time

[Pius]

Homework Statement

A baseball is seen to pass upward by a window 23m above the street with a vertical speed of 14m/s. If the ball was thrown from the street, (a) what was the initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again?

Homework Equations

v=v0+at
x=x0+v0t+1/2at^2
v^2=v02+2a(x-x0)

The Attempt at a Solution

So I guess I use x=x0+v0t+1/2at^2 for the initial speed...altitude and time?

 

[Oscur]

It's a lot clearer if you re-write your equations in a slightly different form:

<br /> v = v_0 + a t<br />

<br /> s = v_0 t + \frac{1}{2} a t^2<br />

<br /> v^2 = v_0^2 + 2 a s<br />

Where s is the displacement x-x_o[/tex](hint hint).

 

[kreil]

You know that at a height of 23m, the ball is moving at 14m/s. From this, you can calculate how much higher the ball will ascend before it begins its descent. The altitude it reaches will be this value plus 23m.

At this moment in time where the ball is still, you can apply freefall mechanics to it in order to find the speed at which it hits the ground. Since the flight of the ball is a parabola, its velocity is symmetric about the turning point, and the speed at which it hits the ground is the speed at which it was thrown (ignoring air resistance).

Once you have this velocity information, you can easily calculate (c) and (d).

 

[ashishsinghal]

hint: the horizontal speed doesn't change