Question

Standing on top of the science building, which is 48 m high, a group of students use a slingshot to shoot a steel ball bearing straight up into the air. The ball bearing passes the students on its way down to the ground 3.2 sec after it was shot. (once the ball bearing leaves the slingshot, it accelerates at -9.8m/s/s)

a) With what speed did the ball bearing leave the slingshot?

b) What was the ball's velocity just before it hit the ground?

Answer 1

a)
Time to reach highest point = 3.2/2=1.6s
0=u-gt
Initial velocity, u=9.8*1.6=?
b)
On its way down, the velocity is -u at the top of the building, and v on the ground. Acceleration is still -9.8 m/s².
Use
v²-u²=2aS
a=-9.8 m/s²
S=-48m
u= calculated above
Solve for v.

For b)
you could also use energy,
(1/2)mv²-(1/2)mu²=mgh
h=48m.