A tennis ball is thrown straight up with an initial speed of 19.5 m/s. It is caught at the same distance above the ground.

(a) How high does the ball rise?

find time of flight:

hf=ho+Vi*t-4.9t^2
solve for t.

Then solve for height at t/2

d(up) = Vo*t + 0.5at^2,

v=at,
a = v/t,
Substitute v/t for a in the 1st Eq:
d(up) = Vo*t + 0.5(v/t)t^2,
d(up) = 19.5t + 0.5 * (19.5)t,

d(down) = 0.5 * 9.8t^2,

d(up) = d(down),
19.5t + 9.75t = 4.9t^2
Solve for t:
29.25t = 4.9t^2,
4.9t^2 - 29.25t = 0,
Factor out t:
t(4.9t - 29.25) = 0,
t = 0,
4.9t - 29.25 = 0,
4.9t = 29.25,
t = 5.97 s.

Solution set: t = 0, and t = 5.97 s.
Select t = 5.97 s.

d(up) = 19.5 m/s * 5.97 s + 9.75 m/s * 5.97 s = 174.6 m.

SI unit of delta V/ delta t and dV/dt

To find out how high the ball rises, we can use the concept of motion in one dimension. We need to find the maximum height reached by the ball.

Here are the steps to solve this problem:

1. Identify the given information:
- Initial velocity (u) = 19.5 m/s
- Final velocity (v) = 0 m/s (when the ball reaches its maximum height)
- Acceleration (a) = -9.8 m/s² (assuming upwards motion as positive)

2. Find the time taken to reach the maximum height:
Use the equation v = u + at, where v = 0, u = 19.5 m/s, and a = -9.8 m/s².
0 = 19.5 - 9.8t
9.8t = 19.5
t = 19.5 / 9.8
t ≈ 2 seconds

3. Calculate the maximum height:
Use the equation s = ut + (1/2)at², where s is the displacement (height) and t is the time taken.
s = 19.5 * 2 + (1/2) * (-9.8) * (2²)
s = 39 + (-9.8) * 2
s = 39 - 19.6
s ≈ 19.4 meters

Therefore, the tennis ball rises to a height of approximately 19.4 meters.