A ball is thrown vertically upwards returns to a boy after 6 seconds ball is 5 M below the highest point 8 equal to 2 second the time at which the ball will be at the same position where gravitational force is equal to 10 M per second square?
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heya man !!!
It is given that the total time of the journey upward and downward is 6 seconds. As the distance travelled by the ball in both the upward and downward journey is the same so, let us take the distance travelled during both upward and downward motion to be ‘d’ meters each. So as per the 1st equation of motion,
v = u+at
In the case of the upward journey,
v = 0 m/s, a = -10 m/s^2 and since the time of both the upward and downward journies is the same so t for upward journey = t for downward journey = 6/2 = 3s.
as per the above mentioned equation
0 = u - 10 x 3
0 = u - 30
u = 30 m/s
Now as per the 2nd equation of motion,
s = ut + 1/2 at^2
s = 30 x 3 + 1/2 x -10 x 9
s = 90 - 45
s = 45 m
So the greatest height reached by the ball is of 45 m and the initial velocity of the ball is 30 m/s.
hope it helps !!
@ himanshu Jha ✌✌
It is given that the total time of the journey upward and downward is 6 seconds. As the distance travelled by the ball in both the upward and downward journey is the same so, let us take the distance travelled during both upward and downward motion to be ‘d’ meters each. So as per the 1st equation of motion,
v = u+at
In the case of the upward journey,
v = 0 m/s, a = -10 m/s^2 and since the time of both the upward and downward journies is the same so t for upward journey = t for downward journey = 6/2 = 3s.
as per the above mentioned equation
0 = u - 10 x 3
0 = u - 30
u = 30 m/s
Now as per the 2nd equation of motion,
s = ut + 1/2 at^2
s = 30 x 3 + 1/2 x -10 x 9
s = 90 - 45
s = 45 m
So the greatest height reached by the ball is of 45 m and the initial velocity of the ball is 30 m/s.
hope it helps !!
@ himanshu Jha ✌✌
chinmayanath143:
Sir here the is asked in seconds and i know that here the answer will be 4seconds
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