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:

$$v = v_0 + a t$$

$$s = v_0 t + \frac{1}{2} a t^2$$

$$v^2 = v_0^2 + 2 a s$$

Where s is the displacement [itex]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

 

[rwisz]

I would approach this problem by first identifying the known variables and the unknown variables. The known variables in this scenario are the vertical speed (14m/s), the altitude (23m), and the acceleration due to gravity (9.8m/s^2). The unknown variables are the initial speed, the time, and the altitude reached by the ball.

To solve for the initial speed, we can use the equation v=v0+at. We know that the vertical speed (v) is 14m/s and the acceleration (a) is -9.8m/s^2 (negative due to the ball moving in the opposite direction of gravity). Therefore, we can rearrange the equation to solve for v0, which gives us an initial speed of 23.8m/s.

To find the altitude reached by the ball, we can use the equation x=x0+v0t+1/2at^2. We know that the initial position (x0) is 0m (since the ball is thrown from the street) and the initial speed (v0) is 23.8m/s. We can also rearrange the equation to solve for t, which gives us a time of 2.44 seconds for the ball to reach the maximum altitude. Substituting this time into the original equation, we can solve for x (altitude), which gives us a maximum altitude of 23.6m.

To find the time at which the ball was thrown, we can use the same equation x=x0+v0t+1/2at^2. We know that the final position (x) is 23m, the initial position (x0) is 0m, and the initial speed (v0) is 23.8m/s. Solving for t gives us a time of 1.48 seconds, which means the ball was thrown 1.48 seconds before it passed the window.

Lastly, to find the time at which the ball reaches the street again, we can use the equation v^2=v02+2a(x-x0). We know that the final speed (v) is 0m/s (since the ball reaches the ground and stops), the initial speed (v0) is 23.8m/s, and the final position (x) is 0m (since the ball reaches the ground). Solving for t gives