Question

20
Numericals in Physics
32. Out of excitement a cricketer throws the ball up in air after catching it
He then recatches it after 4 sec. How below the ball was from its
highest point after 3 sec from start ? (neglect air resistance).
Ans. 4.9 metre,
locity it was thrown.​

Answer 1

Answer:

Explanation:

S = ut + ½at²

Here we are talking about the downfall. So, time taken will be (total time/2) = 2s

So, u = 0 m/s

t = 1 s {As 3 seconds from the start means 1 second from the highest point}

S = (0)(1)+½(9.8)(1)²

= 0+(4.9)

= 4.9 m