A ball is thrown down at 72km h-1 speed from the top of a building. The building is 125 metres tall. The distance travelled before it reached the ground is as follows... $$s = U_0 t + \frac15gt^2$$ where
- Uo = initial velocity (m s-1)
- g = acceleration due to gravity (10m s-2)
- t = time(s)
Find the time for the ball to drop to a fifth of the height of the building.
I have worked out the following but I need it checking, if someone could check it I would appreciate it.
1/5 of the height of the building is 25 meters. So the ball has travelled 100m.
using these equations... x=-b(+or-) Squareroot b^2 -4ac/2a
a=5 b=20 c=-100
Uo=20ms g=10ms S=100m t=20
with the above I have got...
x=-20(+or-) sqroot of 20^2 -4x5-100/2x5
I get the following...
x=-20(+or-) sqroot 2400/10
then taking the positive answer I get
=6.899 AS THE ANSWER.
AND FOR NEGATIVE ANSWER... (-)
20-48.99/10 =-2.899 FOR THE ANSWER.
As the time cannot be negative... the answer is 6.899 secs, but I know I've gone wrong somewhere can anybody clarify this equation for me please?