Question

Question #a3917

Answer 1

Initial upward velocity of the stone thrown `=u=29.8m"/s"`

Retardation acting on the stone due to downward gravitational force is equal to the acceleration due to gravity.
So `g=-9.8m"/"s^2`

i) If stone attains velocity `v` after `t` sec then by equation of kinematics we have

`v=u- g t`

At maximum height the velicity of the stone will be zero.
So `0=29.8-9.8xxt`

`=>t=29.8/9.8~~3.04s`

ii) Again `v^2=u^2-2gh`,where h is height attained at `t` sec

At maximum height `v =0 and h=h_"max"`

So `h_"max"=u^2/(2g)=29.8^2/(2xx9.8)~~45.3m`

iii) And the velocity after 3 sec

`v_3=29.8-9.8*3=0.4m"/s"`