Question

A ball is dropped from a height of 20m above the surface of water in a lake

Answer 1

ANSWER:

• velocity of the ball when it touches the water surface = 20 m/s
• time taken by ball to reach the water surface = 2 seconds

Explanation:

given that,

A ball is dropped from a height of 20m above the surface of water in a lake

here,

since,

the ball was at rest earlier

so,

initial velocity of the ball = 0 m/s

and

given the height from which it was fallen = 20 m

let the velocity at which it touches the water surface be v

now,

we have,

initial velocity(u) = 0 m/s

final velocity(v) = v

height(h) = 20 m

gravitational acceleration(g) = 10 m/s²

by the equation of gravitation,

v² = u² + 2gh

putting the values,

v² = (0)² + 2(10)(20)

v² = 0 + 400

v² = 400

v = √400

v = 20

so,

velocity of the ball when it touches the water surface = 20 m/s

let the time taken by the ball to reach the water surface be t

now,

we have,

initial velocity(u) = 0 m/s

final velocity(v) = 20 m/s

time taken(t) = t

gravitational acceleration(a) = 10 m/s²

by the equation of gravitation

v = u + gt

putting the values,

20 = 0 + 10t

10t = 20

t = 20/10

t = 2

so,

time taken by ball to reach the water surface = 2 seconds