Question

A ball is thrown vertically upwards with a velocity of 20m/s from the top of a multistory building. The height of the point from where the ball is thrown is 25m from the ground. (a) how high will the ball rise? And (b) how long will it be before the ball hits the ground?

Answer 1

Answer:

Initial velocity= 20 m/s

Final velocity= 0 (as it will reach its highest point on air then it will become zero)

Acceleration due to gravity= -9.8 m/s^2 (as the acceleration is not towards gravity that is why it is negative)

Height of the total multistorey building from ground is = 25 m

a. height of the rise of the ball

In order to find height we must first find time.

So, g= v-u/t

=> -9.8 = 0-20/t

=> -9.8 = -20/t (then cross multiply)

=> -9.8t = -20

=> t = -20/-9.8

=> t = 2.04081633 or 2.04 s

h= ut + 1/2 gt^2

= 20 × 2.04 + 1/2 × -9.8 × 2.04 × 2.04

= 40.8 + (-4.9) × 4.1616

= 40.8 + (-20.39184)

= 40.8 - 20.39184

= 20.40816 m

Thus, height is 20.40816 m

b. time

I have already found time as 2.04 s.

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