A ball thrown by one player reaches the other in $$2$$ s. The maximum height attained by the ball above the point of projection will be (Take$$, g $$=$$ $$10$$ $$ms-2$$ $$) [BHU $$2012$$]

Question

A ball thrown by one player reaches the other in $$2$$ s. The maximum height attained by the ball above the point of projection will be (Take$$, g $$=$$ $$10$$ $$ms-2$$ $$)                     [BHU $$2012$$]

Infinity Learn · Accepted Answer

Since, the ball reaches from one player to another in $$2$$ s, so the time period of the flight, T $$=$$ $$2$$ s⇒  $$2u$$ $$sin$$θg $$=$$ $$2$$ sHere, u is the initial velocity and θ is the angle of projection.⇒  u $$sin$$θ $$=$$ g            ......(i)Now$$, we know that the maximum height of the projection,$$H=u2sin2$$θ$$2g$$ or $$H=(usin$$θ$$)22gOn$$ putting the value of u $$sin$$ θ from Eq. $$(i), we $$getH=g22g=g2$$ or $$H=g2=102$$ m or $$H=5$$ m

A ball thrown by one player reaches the other in 2 s. The maximum height attained by the ball above the point of projection will be (Take, g = 10 ms^-2 ) [BHU 2012]

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A ball thrown by one player reaches the other in 2 s. The maximum height attained by the ball above the point of projection will be (Take, g = 10 ms-2 ) [BHU 2012]

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A ball thrown by one player reaches the other in 2 s. The maximum height attained by the ball above the point of projection will be (Take, g = 10 ms^-2 ) [BHU 2012]