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)
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Answer:
6 seconds
Step-by-step explanation:
since the ball moves vertically upwards with an initial velocity = 30m / second
we can apply equation of motion for calculating time.
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→V=u+at
\begin{gathered}→0 = 30 + ( - 9.8)t \\ →9.8t = 30 \\ →t = \frac{30}{9.8} \\ →t = 3.06 \: s = 3 \: s\end{gathered}
→0=30+(−9.8)t
→9.8t=30
→t=
9.8
30
→t=3.06s=3s
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
another method :
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→s=0
Applying Equation of motion,
→ s = ut \: + \frac{1}{2}a {t}^{2}→s=ut+
2
1
at
2
→ 0 = 30t + \frac{1}{2}( - 9.8) {t}^{2}→0=30t+
2
1
(−9.8)t
2
→ (9.8) {t}^{2} = 30t \times 2→(9.8)t
2
=30t×2
→ t = \frac{60}{9.8} = 6.1 \: s→t=
9.8
60
=6.1s
t ≈ 6 seconds.
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