Question

7. A ball is gently dropped from a height of 20 m. If its velocity
increases uniformly at the rate of 10 m sº, 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

v

2

−u

2

=2as

or v

2

−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

Initial velocity[u]=0m/s

Final velocity[v]= v

Height[s]= 20

Acceleration[a]=10

Using,

${v}^{2} = {u}^{2} + 2as$

${v}^{2} = 0 + 2 \times 10 \times 20$

${v }^{2} = 400$

v=20m/s

Hence the ball will strike the ground with speed 20m/s

Now,Using

$s = ut + \frac{1}{2} a {t}^{2}$

$20 = 0 + \frac{1}{2} \times 10 \times {t}^{2}$

T=2sec

Hence After 2sec,the ball will strike the ground