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?

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.

To solve these problems, we can use the equations of motion for an object in free fall. We have the initial height, the time of flight, and the acceleration due to gravity.

a) To find the initial speed of the ball bearing, we can use the equation:

v = u + at

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

We know that the final velocity is 0 since the ball bearing momentarily stops when it reaches the highest point of its trajectory. The acceleration is -9.8 m/s^2 (negative sign because it is directed downwards due to gravity), and the time of flight is 3.2 seconds.

Plugging in these values, we get:

0 = u + (-9.8)(3.2)

Simplifying the equation:

u = 9.8(3.2)

u ≈ 31.36 m/s

Therefore, the ball bearing left the slingshot with a speed of approximately 31.36 m/s.

b) To find the velocity just before the ball bearing hits the ground, we can use the equation:

v = u + at

Here, v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

We know that the initial velocity is 31.36 m/s (as we found in the previous part), the acceleration is -9.8 m/s^2, and the time of flight is 3.2 seconds.

Plugging in these values, we get:

v = 31.36 + (-9.8)(3.2)

Simplifying the equation:

v = 31.36 - 31.36

v = 0

Therefore, the velocity just before the ball bearing hits the ground is 0 m/s (it comes to a complete stop).