Question

In a cricket match, a cricketer tries to play hook shot but the ball takes the edge of his bat and

rises vertically upwards with an initial velocity of 30 m/s. The fielder settles himself under the

ball and takes the catch comfortably after n seconds since the batsman had edged the ball.

What would be the value of n ?

(Take g = 10 m/s2

and neglect any air resistance.
***those who will answer this I will mark him/her as Brainliest***​

Answer 1

Answer : 6 Seconds

Since the ball moves vertically upwards with an initial velocity of 30 m/s,

we can apply Equations of motion for calculating time.

Method I

Here,

u = 30 m/s

V = 0 m/s ( at highest point )

a = -9.8 m/s squared. ( acceleration downwards due to gravity)

Let the time taken be t seconds to the top most point.

We know that,

$→ V = u + at$

$→0 = 30 + ( - 9.8)t \\ →9.8t = 30 \\ →t = \frac{30}{9.8} \\ →t = 3.06 \: s = 3 \: s$

t ≈ 3 seconds

When the ball comes down, it will again take same time, (t = 3 seconds).

Hence,

Total time (n) after which fielder catches the ball

= 6 seconds.

Method II

Since the ball has only vertical velocity, it will come back to the same position after falling down.

Hence,

The final displacement of the ball = 0

$→ s = 0$

Applying Equation of motion,

$→ s = ut \: + \frac{1}{2}a {t}^{2}$

$→ 0 = 30t + \frac{1}{2}( - 9.8) {t}^{2}$

$→ (9.8) {t}^{2} = 30t \times 2$

$→ t = \frac{60}{9.8} = 6.1 \: s$

t ≈ 6 seconds.

Hope this helps you and makes your concepts clear :)

Answer 2

6 Seconds

EXPLAIN :-

Since the ball moves vertically upwards with an initial velocity of 30 m/s, we can apply Equations of motion for calculating time.