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(a) Initial velocity of ball (u)=50m/s, acceleration of ball =−g,
final velocity at the highest point (v)=0
So applying the 3rd equation of motion we get:
v2=u2−2ghmax
⇒0=502−2×10×hmax
⇒hmax=202500=125m
(b) Let the time required to reach max height be t. Then applying 1st equation of motion we get:
v=u−gt
⇒0=50−10t
⇒t=5s
(c) Let speed at half of max height be V then:
V2=502−2g2125
⇒V2=2500−1250=1250
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