1. A boy is standing at a distance a1 from the foot of a tower.The boy throws an stone at a angle 45° 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°.Intial velocty of the ball is u0 m/s and at an angle 65°(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 θ 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 θ,R and v0
4. The displacement of the body x(in meters) varies with time t (in sec) as
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?
3. The time taken to reach the pole is given
Now equation of motion of vertical motion
at t=R/v0cosθ h=H
H=(v0sinθ)X(R/v0cosθ ) -(1/2)g( R/v0cosθ)2
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)
then t=4/3 sec
Displacement can be found by substituing the value of t=4/3 in equation (1)
Class 9 physics notes on chapter motion is now available at the following link CLASS 9 MOTION NOTES (FULL LENGTH)
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