kinematics questions with solutions

In this article find kinematics questions with solutions to some selected problems.

Kinematics Objective type questions

Kinematics Numericals

1. A boy is standing at a distance a1 from the foot of a tower.The boy throws an stone at a angle 45&#176 which just touches the top of the tower and strikes the ground at a distance a2 from the point the boy is standing.Find the height of the tower

2. A ball is thrown upward from a point on the side of a hill which slopes upward uniformly at angle 45&#176.Intial velocty of the ball is u0 m/s and at an angle 65&#176(with respect to the horizotal).At what distance up the slope the ball strike and in what time?

3.A cannon on a level plain is aimed at an angle &#952 above the horizontal.A shell is fired with a muzzle velocity v0 towards a pole which is distance R away.It hits the pole at height H.
a find the timetaken to reach the pole
b. find the value of H in terms of &#952,R and v0

4. The displacement of the body x(in meters) varies with time t (in sec) as
x=(-2/3)t2 +16t+2
find following
a. what is the velocty at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d .what will the displacement when it comes to rest
e .How much time it take to come to rest.

5. A man runs at a speed at 4 m/s to overtake a standing bus.When he is 6 m behind the door at t=0,the bus moves forward and continues with constant acceleration of 1.2 m/s2
find the following
a. how long does it take for the man to gain the door
b if in the beginning he is 10m behind the door ,will he running at the same speed ever catch up bus?

Detailed Solutions:
3. The time taken to reach the pole is given
R=v0cos&#952 t
or t=R/v0cos&#952

Now equation of motion of vertical motion
at t=R/v0cos&#952 h=H
H=(v0sin&#952)X(R/v0cos&#952 ) -(1/2)g( R/v0cos&#952)2
or H=Rtan&#952-gR2/2v02cos2&#952


Given x=(-2/3)t2 +16t+2

so velocity at t=0 =16
and velocity at t=1 =(-4/3)+16=44/3

Acceleration is given as =d2x/dt2=-4/3

so acceleration is time independent and it is constant

Displacement at t=0 can be found by simply substituting the values of t=0 in equation (1)

Now v=(-4/3)t+16
when v=0
then t=4/3 sec

Displacement can be found by substituing the value of t=4/3 in equation (1)
=(-32+576+54)/27=598/27 m

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