Question #a3917

1 Answer
Jan 25, 2017

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"#