Answer:

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Explanation:

Height of tower =60 m

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 af after 4 seconds going downward.

Again using, s=ut+1/2gt^2 we get,

60=20t+1/2×10t^2

Or, 60=20t+5t^2

Or,t^2+4t-12=0

Or,t^2+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=2 seconds

Therefore, time for the ball to reach the ground after release is:(4+2)=6 seconds.

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Explanation:

height of tower=60 m

initial velocity = 20m/s

final velocity = 0

using, v=u -gt

or ,0 = 20 - 10t

or,10t = 20

or,t = 25

Now, time of ascent = time of descent

Therefore, the ball pass the man after 4 seconds going downward.

Again using ,s =ut+1/2gt^2 we get,

60=20t + 1/20t ×10 +12

Or, 60= 20t +5+72

or ,+ 72t 4t - 12 = 0

Either, t=6 which is not possible or, t=2 seconds .

Therefore , time for the ball to reach the ground after release is :(4+2)=6 seconds ....

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