Question #a3917

Question

1689 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-25T14:53:03

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

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