Question

A ball is gently dropped from a height of 20 m. If its velocity Increases uniformly at the rate of 10 m s2, with what velocity will it strike the ground? After what time will it strike the ground?​

Answer 1

20 m/s

Step-by-step explanation:

Height (s)= 20m

acceleration (a) = 10 m/s^2

v=?

we know,

V^2= U^2 + 2as. (U is initial velocity)

or,. V^2 = 0 + 2 × 10 × 20

or, V^2 = 400

or,. V = 20 m/s

Answer 2

Step-by-step explanation:

Initial Velocity of the ball (u) = 0

Distance or height of fall (s) = 20 m

Downward acceleration (a) = 10 m s-2

As we know

2as = v2 – u2

v2 = 2as + u2

v2 = (2 x 10 x 20 ) + 0

v2 = 400

Final velocity of ball (v) = 20 ms-1

t = (v – u)/a

Time taken by the ball to strike (t) = (20 – 0)/10

t = 20/10

t = 2 seconds

The final velocity with which the ball will strike the ground is (v) = 20 ms-1

The time it takes to strike the ground (t) = 2 seconds