A body is released to fall under the influence of gravity and its equation of motion is given by `(t)=(9t^(2)+4)sec`.Find the ratio of the instantaneous velocities of the body at the fifth second and seventh second after release: A. `1:2` B. `5:7` C. `5:3` D. `1:11`

A body is released to fall under the influence of gravity and its equation of motion is given by s(t)=(9t^(2)+4)sec .Find the ratio of the instantaneous velocities of the body at the fifth second to that at the seventh second after release. A). 2:9 B). 1:2 C). 1:11 D). 5:7

The displacement of a body at any time t after starting is given by s=15t-0.4t^2 . Find the time when the velocity of the body will be 7ms^-1 .

The displacement of a body is given by s=(1)/(2)g t^(2) where g is acceleration due to gravity. The velocity of the body at any time t is

The velocity of a particle is given by v=(2t^(2)-3t+10)ms^(-1) . Find the instantaneous acceleration at t = 5 s.

A body is projected in a vertical direction. The altitude y is given by y=10t^(2)-9t+5 Calculate the initial velocity of the body.

A body is moving vertically upwards under gravity such that its position from ground is given as y=ut-(1)/(2)g t^(2) . Find the max height reached by body.

A body of mass 10 kg is upon by a force given by equation F=(3t^(2)-30) newtons. The initial velocity of the body is 10 m/s. The velocity of the body after 5 sec. is

Equation of motion of a body is (dv)/(dt) = -4v + 8, where v is the velocity in ms^-1 and t is the time in second. Initial velocity of the particle was zero. Then,

A ball is released from the top of a tower. The ratio of work done by force of gravity in 1st second, 2nd second and 3rd second of the motion of ball is

The velocity of a particle is given by v=(2t^(2)-4t+3)m//s where t is time in seconds. Find its acceleration at t=2 second.