A person throws balls into air vertically upward in regular intervals of time of one second . The next ball is thrown when the velocity of the ball thrown earlier becomes zero . The height to which the balls rise is …
(Assume , g = `10 ms^(-2)`)

Options

A. 5 m
B. 10 m
C. 7.5 m
D. 20 m

Answer

A

Explanation

Time taken by the ball to reach highest point is t = 1 s
As the person throws the second ball , when the velocity of the first ball becomes zero, i.e, v =0 or when the first ball reach the highest point ,
Using v = u + at
Here , v =0 , a = -g , t = 1 s
`therefore 0 = u - (10) (1)`
Using `v^(2) - u^(2) = 2ah` , we get
`(0)^(2) - (10)^(2) = 2 (-10) (h)`
`h = ((10)^(2))/(20) = 5` m