An elevator without a ceiling is ascending up with an acceleration of $$5$$ m $$s-2$$. A boy on the elevator shoots a ball in vertically upward direction from a height of $$2$$ m above the floor of elevator. At this instant, the elevator is moving up with a velocity of $$10$$ m $$s-1$$ and floor of the elevator is at a height of $$50$$ m from the ground. The initial speed of the ball is $$15$$ m $$s-1$$ w.r.t. the elevator. Consider the duration for which the ball strikes the floor of the elevator in answering following questions:The time in which the ball strikes the floor of elevator is given by The maximum height reached by the ball as measured from the ground would beThe displacement of ball w.r.t. ground during its flight is

Question

An elevator without a ceiling is ascending up with an acceleration of $$5$$ m $$s-2$$. A boy on the elevator shoots a ball in vertically upward direction from a height of $$2$$ m above the floor of elevator. At this instant, the elevator is moving up with a velocity of $$10$$ m $$s-1$$ and floor of the elevator is at a height of $$50$$ m from the ground. The initial speed of the ball is $$15$$ m $$s-1$$ w.r.t. the elevator. Consider the duration for which the ball strikes the floor of the elevator in answering following questions:The time in which the ball strikes the floor of elevator is given by The maximum height reached by the ball as measured from the ground would beThe displacement of ball w.r.t. ground during its flight is

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Initial velocity of ball w.r.t. ground.v→$$BG=v$$→$$BE+v$$→$$EG=15+10=25ms$$−$$1$$↑v→$$BE=15ms$$−$$1$$↑,a→$$BG=10ms$$−$$2$$↓a→$$BE=a$$→BG−a→$$EG=10$$−$$($$−$$5)=15ms$$−$$2$$↓s→$$BE=2m$$↓. Using $$s=ut+12at2w$$.r.t. elevator frame,−$$2=15t$$−$$12$$×$$15t2$$⇒$$t=2.13sThe$$ height of the ball from ground at the time of projection is $$52$$ m. Maximum height from point of projection $$ish=v$$→$$BG22aBG=(25)2$$×$$10=31.25mRequired$$ height $$=$$ $$52$$ $$+$$ $$31.25$$ $$=$$ $$83.25$$ mDisplacement of floor of elevator in $$2.13$$ s $$iss1=10t+12$$×$$5$$×$$t2=32.64mSo$$ displacement of ball w.r.t. ground, s→$$BG=s$$→$$BE+s$$→$$EG=$$−$$2+32.64=30.64m$$

Rectilinear Motion

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The time in which the ball strikes the floor of elevator is given by

The maximum height reached by the ball as measured from the ground would be

The height of the ball from ground at the time of projection is 52 m. Maximum height from point of projection is
$h = \frac{{|{\vec{v}}_{BG}|}^{2}}{2 a_{BG}} = \frac{(25)}{2 \times 10} = 31 .25 m$
Required height = 52 + 31.25 = 83.25 m

The displacement of ball w.r.t. ground during its flight is

Displacement of floor of elevator in 2.13 s is
$s_{1} = 10 t + \frac{1}{2} \times 5 \times t^{2} = 32 .64 m$
So displacement of ball w.r.t. ground,
${\vec{s}}_{BG} = {\vec{s}}_{BE} + {\vec{s}}_{EG} = - 2 + 32 .64 = 30 .64 m$

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