A golfer standing on level ground hits a ball with a velocity of u $$=$$ $$52$$ $$ms-1$$ at an angle α above the horizontal. If $$tan$$ α $$=$$ $$5/12$$, then the time for which the ball is at least $$15$$ m above the ground will be (take$$ g $$=$$ $$10$$ $$ms-2)

Question

A golfer standing on level ground hits a ball with a velocity of u $$=$$ $$52$$ $$ms-1$$ at an angle α above the horizontal. If $$tan$$ α $$=$$ $$5/12$$, then the time for which the ball is at least $$15$$ m above the ground will be (take$$ g $$=$$ $$10$$ $$ms-2)

Infinity Learn · Accepted Answer

Let at any time t, the ball be at height of $$15$$ m.$$Sy=uyt+12ayt2$$ $$15=usin$$α$$t-12gt2$$ $$15=52$$×$$513t-12$$×$$10t2$$ $$t2-4t+3=0$$  (t$$ $$-$$ $$1)(t$$ $$-$$ $$3) $$=$$ $$0$$   t $$=$$ $$1$$ s, t $$=$$ $$3$$ s. Required time is $$3$$ $$-$$ $$1$$ $$=$$ $$2$$ s

A golfer standing on level ground hits a ball with a velocity of u = 52 ms^-1 at an angle $α$ above the horizontal. If tan $α$ = 5/12, then the time for which the ball is at least 15 m above the ground will be (take g = 10 ms-²)

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A golfer standing on level ground hits a ball with a velocity of u = 52 ms-1 at an angle α above the horizontal. If tan α = 5/12, then the time for which the ball is at least 15 m above the ground will be (take g = 10 ms-2)

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A golfer standing on level ground hits a ball with a velocity of u = 52 ms^-1 at an angle $α$ above the horizontal. If tan $α$ = 5/12, then the time for which the ball is at least 15 m above the ground will be (take g = 10 ms-²)