Soln: The velocity of the body when it touches the ground sliding down an inclined plane ,will be same as when the body vertically falls freely from height ‘h”.
From the problem height S=h=5m; acceleration due to gravity g=9.8; Initial velocity u=0 ; V=?
substitute the values in the equation = 2gs ,
we get = 2(9.80) (5) ; = 98
V =9.899 m/sec.
7.A ball projected vertically upwards returns to ground after 15sec.Calculate i)maximum height to which it rises ii)velocity with which it is projected and iii)its position after 6 seconds.
Soln:From the data given in the problem,
Velocity of projection u=?
acceleration due to gravity g=9.8,
time of flight T =15 sec
substitute the values of T,g in the equation T =
=15 ; 2u = 15 (9.80) = 147
ii )Velocity with which it is projected u= 73.5 m/sec.
i)Maximum height =,
substitute the values of u,g in the equation we get =
= =275.625 m
1iii) Let Its position after t=6sec be S
substitute the values of u,t and g in the equation S=ut -
S = (73.5)(6) - =441-176.4 = 264.6 m high from the ground.
8.A stone was dropped from a rising baloon at a height of 150m above the ground and it reaches the ground in 15 seconds.Find the velocity of the baloon at the instant the stone was dropped.
Soln: From the data given in the problem,
Acceleration due to gravity g=9.8 ,
Height of the balloon when the stone is dropped from it h=150m,
Time of flight T=15 sec,
Let the velocity of the baloon when the stone is dropped from it is u (say).
Substitute the values of g,t and h in the equation h = -ut
we get 150 = -u(15) = (4.9)(225) - 15u,
150 = 15[(4.9)(15)- u],
10=73.5 - u ; u = 73.5- 10 =63.5 m/sec.
9. From the top of a tower of 200 metres high, a stone is projected vertically upwards with a velocity 49m/sec.Calculate the i)maximum height traveled by it from ground level, ii)the velocity with which it strikes the ground and the iii) time it takes to reach the ground.
Soln: From the data given in the problem,Height of the tower is h=200m,
velocity of projection u = 49 m/sec,
Max height of the stone from top of the tower be S= X (say),
velocity at maximum height v = 0,
i) substitute the values of u,v and g in equation = -2gs,
we get = -2(9.8)X,
X = = 122.5 m from the top of the tower
Maximum height from ground level H = X +h =122.5 + 200 = 322.5 m.
ii) Let the velocity with which it reaches the ground be say,
substitute the values in the equation =
= 79.50 m/sec.
iii) Time of flight T =?
Substitute the values u,g and h in the equation h = -ut,
we get 200 = -49T,
200 = - 49T
- 10T = ,
by solving this quadratic equation for T we get T= 13.11 sec