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Question Number 17816 by Tinkutara last updated on 11/Jul/17
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
$$\mathrm{A}\:\mathrm{man}\:\mathrm{is}\:\mathrm{standing}\:\mathrm{on}\:\mathrm{top}\:\mathrm{of}\:\mathrm{a}\:\mathrm{building} \\ $$$$\mathrm{100}\:\mathrm{m}\:\mathrm{high}.\:\mathrm{He}\:\mathrm{throws}\:\mathrm{two}\:\mathrm{balls} \\ $$$$\mathrm{vertically},\:\mathrm{one}\:\mathrm{at}\:{t}\:=\:\mathrm{0}\:\mathrm{and}\:\mathrm{other}\:\mathrm{after} \\ $$$$\mathrm{a}\:\mathrm{time}\:\mathrm{interval}\:\left(\mathrm{less}\:\mathrm{than}\:\mathrm{2}\:\mathrm{seconds}\right). \\ $$$$\mathrm{The}\:\mathrm{later}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{at}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of} \\ $$$$\mathrm{half}\:\mathrm{the}\:\mathrm{first}.\:\mathrm{The}\:\mathrm{vertical}\:\mathrm{gap}\:\mathrm{between} \\ $$$$\mathrm{first}\:\mathrm{and}\:\mathrm{second}\:\mathrm{ball}\:\mathrm{is}\:+\mathrm{15}\:\mathrm{m}\:\mathrm{at}\:{t}\:=\:\mathrm{2}\:\mathrm{s}. \\ $$$$\mathrm{The}\:\mathrm{gap}\:\mathrm{is}\:\mathrm{found}\:\mathrm{to}\:\mathrm{remain}\:\mathrm{constant}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{with}\:\mathrm{which}\:\mathrm{the} \\ $$$$\mathrm{balls}\:\mathrm{were}\:\mathrm{thrown}\:\mathrm{and}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{time} \\ $$$$\mathrm{interval}\:\mathrm{between}\:\mathrm{their}\:\mathrm{throw}. \\ $$
Answered by mrW1 last updated on 11/Jul/17
h_1 =v_1 t+(1/2)gt^2 h_2 =(1/2)v_1 (t−Δt)+(1/2)g(t−Δt)^2 Δh=h_1 −h_2 =(1/2)v_1 (t+Δt)+(1/2)g(2t−Δt)Δt =(1/2)v_1 t+gtΔt+(1/2)v_1 Δt−(1/2)gΔt^2 =t((1/2)v_1 +gΔt)+(1/2)Δt(v_1 −gΔt) since Δh remains constant ⇒(1/2)v_1 +gΔt=0 ⇒v_1 =−2gΔt Δh=(1/2)Δt(v_1 −gΔt)=−(3/2)gΔt −15=−(3/2)gΔt ⇒Δt=1 s ⇒v_1 =−2g×1=−20 m/s (⇈)
$$\mathrm{h}_{\mathrm{1}} =\mathrm{v}_{\mathrm{1}} \mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \\ $$$$\mathrm{h}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}_{\mathrm{1}} \left(\mathrm{t}−\Delta\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}−\Delta\mathrm{t}\right)^{\mathrm{2}} \\ $$$$\Delta\mathrm{h}=\mathrm{h}_{\mathrm{1}} −\mathrm{h}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}_{\mathrm{1}} \left(\mathrm{t}+\Delta\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{2t}−\Delta\mathrm{t}\right)\Delta\mathrm{t} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}_{\mathrm{1}} \mathrm{t}+\mathrm{gt}\Delta\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}_{\mathrm{1}} \Delta\mathrm{t}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\Delta\mathrm{t}^{\mathrm{2}} \\ $$$$=\mathrm{t}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}_{\mathrm{1}} +\mathrm{g}\Delta\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{2}}\Delta\mathrm{t}\left(\mathrm{v}_{\mathrm{1}} −\mathrm{g}\Delta\mathrm{t}\right) \\ $$$$\mathrm{since}\:\Delta\mathrm{h}\:\mathrm{remains}\:\mathrm{constant} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\mathrm{v}_{\mathrm{1}} +\mathrm{g}\Delta\mathrm{t}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{v}_{\mathrm{1}} =−\mathrm{2g}\Delta\mathrm{t} \\ $$$$\Delta\mathrm{h}=\frac{\mathrm{1}}{\mathrm{2}}\Delta\mathrm{t}\left(\mathrm{v}_{\mathrm{1}} −\mathrm{g}\Delta\mathrm{t}\right)=−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{g}\Delta\mathrm{t} \\ $$$$−\mathrm{15}=−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{g}\Delta\mathrm{t} \\ $$$$\Rightarrow\Delta\mathrm{t}=\mathrm{1}\:\mathrm{s} \\ $$$$\Rightarrow\mathrm{v}_{\mathrm{1}} =−\mathrm{2g}×\mathrm{1}=−\mathrm{20}\:\mathrm{m}/\mathrm{s}\:\left(\upuparrows\right) \\ $$
Commented by Tinkutara last updated on 11/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 11/Jul/17
Assuming, balls are thrown vertically upwards. First ball at a velocity 2u, and second after time t_0 at velocity u. if t>t_0 we have s_1 −s_2 =15 (s_1 −s_2 =−15 not possible) [2ut−(1/2)gt^2 ]−[u(t−t_0 )−(1/2)g(t−t_0 )^2 ]=15 ⇒ u(t+t_0 )−(g/2)t_0 (2t−t_0 )−15=0 since this is true for all times t>t_0 until any ball hits the ground, so coefficient of t in equation and the constant term are both zero. u−gt_0 =0 and ut_0 +((gt_0 ^2 )/2)−15=0 or u=gt_0 and substituting this in second equation, we get ((3gt_0 ^2 )/2)=15 ⇒ t_0 =(√((10)/g)) u=gt_0 =(√(10g)) if g=10m/s^2 , t_0 =1s and u=10m/s.
$$\mathrm{Assuming},\:\mathrm{balls}\:\mathrm{are}\:\mathrm{thrown}\: \\ $$$$\mathrm{vertically}\:\mathrm{upwards}.\:\mathrm{First}\:\mathrm{ball}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{velocity}\:\mathrm{2u},\:\mathrm{and}\:\mathrm{second}\:\mathrm{after}\:\mathrm{time} \\ $$$$\:\mathrm{t}_{\mathrm{0}} \:\mathrm{at}\:\mathrm{velocity}\:\mathrm{u}. \\ $$$$\mathrm{if}\:\:\mathrm{t}>\mathrm{t}_{\mathrm{0}} \:\mathrm{we}\:\mathrm{have} \\ $$$$\:\mathrm{s}_{\mathrm{1}} −\mathrm{s}_{\mathrm{2}} =\mathrm{15}\:\:\:\:\:\left(\mathrm{s}_{\mathrm{1}} −\mathrm{s}_{\mathrm{2}} =−\mathrm{15}\:\mathrm{not}\:\mathrm{possible}\right) \\ $$$$\:\left[\mathrm{2ut}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \right]−\left[\mathrm{u}\left(\mathrm{t}−\mathrm{t}_{\mathrm{0}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\mathrm{t}−\mathrm{t}_{\mathrm{0}} \right)^{\mathrm{2}} \right]=\mathrm{15} \\ $$$$\Rightarrow\:\:\:\mathrm{u}\left(\mathrm{t}+\mathrm{t}_{\mathrm{0}} \right)−\frac{\mathrm{g}}{\mathrm{2}}\mathrm{t}_{\mathrm{0}} \left(\mathrm{2t}−\mathrm{t}_{\mathrm{0}} \right)−\mathrm{15}=\mathrm{0} \\ $$$$\:\mathrm{since}\:\mathrm{this}\:\mathrm{is}\:\mathrm{true}\:\:\mathrm{for}\:\mathrm{all}\:\mathrm{times}\:\mathrm{t}>\mathrm{t}_{\mathrm{0}} \\ $$$$\mathrm{until}\:\mathrm{any}\:\mathrm{ball}\:\mathrm{hits}\:\mathrm{the}\:\mathrm{ground},\:\mathrm{so} \\ $$$$\:\mathrm{coefficient}\:\mathrm{of}\:\boldsymbol{\mathrm{t}}\:\mathrm{in}\:\mathrm{equation}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{constant}\:\mathrm{term}\:\mathrm{are}\:\mathrm{both}\:\mathrm{zero}. \\ $$$$\:\mathrm{u}−\mathrm{gt}_{\mathrm{0}} =\mathrm{0}\:\:\:\:\mathrm{and}\:\:\:\:\mathrm{ut}_{\mathrm{0}} +\frac{\mathrm{gt}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}−\mathrm{15}=\mathrm{0} \\ $$$$\:\:\mathrm{or}\:\:\mathrm{u}=\mathrm{gt}_{\mathrm{0}} \:\mathrm{and}\:\mathrm{substituting}\:\mathrm{this}\:\mathrm{in} \\ $$$$\mathrm{second}\:\mathrm{equation},\:\mathrm{we}\:\mathrm{get} \\ $$$$\:\:\:\:\frac{\mathrm{3gt}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}=\mathrm{15}\:\:\:\Rightarrow\:\:\:\:\mathrm{t}_{\mathrm{0}} =\sqrt{\frac{\mathrm{10}}{\mathrm{g}}}\: \\ $$$$\:\:\:\:\:\mathrm{u}=\mathrm{gt}_{\mathrm{0}} \:=\sqrt{\mathrm{10g}} \\ $$$$\mathrm{if}\:\:\mathrm{g}=\mathrm{10m}/\mathrm{s}^{\mathrm{2}} ,\:\mathrm{t}_{\mathrm{0}} =\mathrm{1s}\:\mathrm{and}\:\mathrm{u}=\mathrm{10m}/\mathrm{s}. \\ $$
Commented by Tinkutara last updated on 11/Jul/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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