Question

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?

Answer 1

find time of flight:
hf=ho+Vi*t-4.9t^2
solve for t.

Then solve for height at t/2

Answer 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.

Answer 3

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