a man is standing at the top of a 60m high tower. He throws a ball vertically upwards with velocity 20m/s. After what time will the ball pass him going downward?How long after it release will the ball reach the ground?(take g=10m/s square)
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Answered by
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Height of the tower = 60m
Initial velocity = 20m/s
Final velocity = 0
Using, v=u-gt
or, 0=20-10t
or, 10t=20
or, t=2s
now, time of ascent= time of descent
therefore, the ball pass the man after 4s going downward.
againg using, s=ut+1/2gt² we get,
60=20t+1/2×10t²
or, 60=20t+5t²
or, t²+4t-12=0
or, t²+6t-2t-12=0
or, t(t+6)-2(t+6)=0
or, (t+6)(t-2)=0
either, t=-6 which is not possible
or, t=2s
∴, time for the ball to reach the ground after release is:(4+2)=6s
Initial velocity = 20m/s
Final velocity = 0
Using, v=u-gt
or, 0=20-10t
or, 10t=20
or, t=2s
now, time of ascent= time of descent
therefore, the ball pass the man after 4s going downward.
againg using, s=ut+1/2gt² we get,
60=20t+1/2×10t²
or, 60=20t+5t²
or, t²+4t-12=0
or, t²+6t-2t-12=0
or, t(t+6)-2(t+6)=0
or, (t+6)(t-2)=0
either, t=-6 which is not possible
or, t=2s
∴, time for the ball to reach the ground after release is:(4+2)=6s
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97
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