A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
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1 Answer
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This is a long answer type question as classified in NCERT Exemplar
let speed of two balls be V1and V2
Where v1=2v and v2=v and y1and y2 be the distance covered
So y1= and y2=
So y1-y2= 15
V2=
So clearly we can say v1=20 and v2=10
And y1=20m and y2=5m
If t2 is the time taken by ball 2 through a distance of 5m, y2=v2t-1/2gt2
5=10t2-5t22 so t2 will be 15
Then time covered by ball 1 in 2 sec between two throws = t1-t2= 2-1=1s
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Please find the solution below:
[h] = ML2T-1
[E] = ML2T-2
[V] = ML2T-2C-1
[P] = MLT-1
According to question, we can write
10 =
Average speed
(d) Initial velocity
Final velocity
Change in velocity
Momentum gain is along
Force experienced is along
Force experienced is in North-East direction.
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