Two equations of motion are: v = u +at (v- final velocity, u- initial velocity, and a acceleration, and

s = ut + 0.5at^2 (s- distance travelled)

Since the stone is thrown up this becomes: 0 = u – 9.8t. or

t = u/9.8 m/sec.

This is because at the end of time t the rock stops rising and starts to fall back. Assuming there is no air resistance the rock will take the same time t to come back to the starting point at the top of the building. It will have also the same velocity u but falling down.

Since the total time was 4 secs., the time to reach the ground from the top of the building will be = (4-2t) = (4 – 2 x (u/9.8)) = 4 – (u/4.9) secs.

Applying the second equation:

u x (4 – (u/4.9)) + 4.9 x (4 – (u/4.9))^2 = 55.8

After simplifying this it becomes: 4u = 16 x 4.9 – 55.8

From this: u = 5.65 m/sec

t = u/9.8 = 5.65/9.8 = 0.577 secs.

Height travelled s = 5.65 x 0.577 – 4.9 x 0.577^2

= 1.63 m

Verify by checking falling distance:

falling time = 4 – (5.65/4.9) = 2.85 secs., u = 5.65 m/sec

So, distance s = 5.65 x 2.85 + 4.9 x 2.85^2 = 55.8 m

This matches the height of he building. So the solution is correct.

Initial velocity = 5.65 m/sec

Height travelled above building = 1.63 m

Keep the dollar to yourself and we are just humans (almost).