Finding Time?

Question

a ball is thrown up at speed 12m/s and at height h how do i find the time at the maximum height and when it reaches the ground? also information is that after the ball reaches the ground its velocity was -30m/s and the time was 4.4s. Thanks

Answer 1

a = -9.8
v = -9.8t + 12
y = -4.9t² + 12t + h

We know that y(t=4.4) = 0:
0 = -4.9(4.4²) + 12(4.4) + h
Solve for h:
h = 4.9(4.4²) - 12(4.4) = 42.064 m
The ball was thrown from a height of 42.064 m.

Maximum height occurs when v = 0:
0 = -9.8t + 12
t = 12/9.8 = 1.22 sec.

You already gave the time when it reaches the ground; why are you asking for it?

Answer 2

When it is thrown from a building of height h,

The initial velocity in the upward direction is U = 12m/s.
The final velocity when it reaches the maximum height is V = 0.
The acceleration due to gravity which acts down ward a = - 9.8m/s^2.

Using V = U + a t, we can find t.

t = 12 / 9.8 = 1.22 second.

For the second part of your question,

The initial velocity in the upward direction is U = 12m/s.
The final velocity when it reaches the ground is V= minus 30m/s ( since the velocity is down ward)
The acceleration due to gravity which acts down ward a = - 9.8m/s^2.

Using V = U + a t, we can find t.
t = (-30 - 12) / (-9.8) = 4.29 second.


h can be found either using V^2 - U^2 = 2as or S = ut + 0.5 a tt.