Question

A ball is thrown vertically upwards with a velocity of 20ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0m from the ground.
(a) How high will the ball rise?
(b) how long will it be before the ball hits the ground?

Give me in detail, step by step please..

Answer 1

(a)
H = u²/(2g)
H = 20²/(2×10)
H = 20 m

The ball will reach 20 m high from the point of projection.
It reaches 20 + 20.5 = 40.5m high from the ground.

(b)
t1 = 2u/g
t1 = 2×20/10
t1 = 4 seconds

If v is the velocity with which it hits the ground then,
v = √(u² + 2aS)
v = √(20² + 2×10×25)
v = 30 m/s

v = u + at
t2 = (v - u)/g
t2 = (30 - 20)/10
t2 = 1 second

Time taken for it to hit the ground = t1 + t2
= 4 + 1
= 5 seconds

Note: I took g = 10m/s².

Answer 2

Answer:

Explanation:

Solution,

(a) Here,

u = + 20 m/s

g = - 10 m/s²

At the highest point, v = 0

Suppose the balls rise to the highest h from the point of projection.

v² - u² = 2gs

⇒ (0)² - (20)² = 2 × (- 10) × h

⇒ h = + 20 m.

Hence, the ball rises to 20 m.

(ii) Net displacement, s = - 25 m.

Negative sign is taken because displacement is in the opposite direction of the initial velocity.

s = ut + 12 gt²

⇒ - 25 = 20t + 1/2 × (- 10) × t²

⇒ 5t² - 20t - 25 = 0

⇒ t² - 4t - 5 = 0

⇒ (t + 1) (t - 5) = 0

⇒ t ≠ - 1, 5 seconds

⇒ t = 5 seconds.

Hence, the ball hits the ground to 5 seconds.