A ball is thrown vertically upward from the ground at time t=0 s. It passes the top of a tower at t=3 s and 2 s later it reaches and its maximum height. The height of the tower is (Acceleration due to gravity, `g=10m//s^(2))`

Options

A. 105 m
B. 125 m
C. 85 m
D. 65 m

Answer

D

Explanation

According to question, ball reaches at top of a tower at the time `t_(1)=3s` and further reaches at the maximum height at time `t_(2)=2s.`
`:.` Total time taken by ball to reached the maximum height,
`t=t_(1)+t_(2)=3+2`
`:.t=5s`
If u be velocity of ball at time, t=0 then from the first equation of the motion,
`v=u-g""t" "[:'g=10m//s^(2)]`
`0=u-10xx5impliesu=50m//s`
If h be the height of the tower, then from the second equation of the motion,
`h=ut-(1)/(2)xxg""t^(2)=50xx3-(1)/(2)xx10xx3^(2)`
`=150-45=105m`
Hence, the height of the tower is 105 m.