A ball 4 s after the instant it was thrown from the ground passes through a point P, and strikes the ground after 5 s from the instant it passes through the point P. Assuming acceleration due to gravity to be `9.8 m//s^(2)` find height of the point P above the ground.

Explanation

The ball projected with velocity `vec(u)=u_(x)hat(i)+u_(y)hat(j)` from O reachrs the point P with velocity `vec(v)=u_(x)hat(i)+v_(y)hat(j)` and hits the ground at point Q at the instant `T=4+5=9s` as shown in the adjoining motion diagram.

From equation of time of flight, we have its initial y-component of velocity `u_(y)`
`T=(2u_(y))/g rarr u_(y)=1/2 g T`
Substituting above in eq. (ii) and rearranging terms, we have the height y of the point P.
`y=u_(y)t-1/2g t^(2) rarr y=1/2g t(T-t)=1/2xx9.8xx4(9-4)=98m`