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

Explanation:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms^−2

Time of fall, t=?

We know,

v²−u²=2as

or v²−0=2× 10 × 20 = 400 or v = 20ms^−1

Now using v =u + at

we have

20=0 + 10 × t or t = 2s

Answer 2

Answer:

(｡◕‿◕｡)➜2 seconds

Explanation:

Let us assume, the final velocity with which ball will strike the ground be ‘v’ and time it takes to strike the ground be ‘t’.

☞Given parameters

☞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 ball will strike the ground is (v) = 20 ms-1

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