Question

A stone projected so as to reach a height

of 72 m. If velocity of projection is v then

velocity of stone at point P And Q is

⁄2and ⁄3, then distance between points

P and Q will be :​

Answer 1

Answer:

At P,(2u)2=u2−2gs1−(1)

At Q,(3u)2=u2−2gs2−(2)

Maximum height =2gu2−(3)

eqn(2)−eqn(1)⇒−2g(s2−s1)=9u2−4u2

⇒+2gh=365u2

⇒536h=2gu2→ from eqn(3)

Max ht =536h

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