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Question Number 153041 by nadovic last updated on 04/Sep/21

A ball is thrown vertically upwards with a velocity of 10 ms^(−1) from a point 75m above the ground. How long will it take the ball to strike the ground?

$$\mathrm{A}\:\mathrm{ball}\:\mathrm{is}\:\mathrm{thrown}\:\mathrm{vertically}\:\mathrm{upwards} \\ $$$$\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{10}\:{ms}^{−\mathrm{1}} \:\mathrm{from}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{75}{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}.\:\mathrm{How}\:\mathrm{long}\:\mathrm{will} \\ $$$$\mathrm{it}\:\mathrm{take}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{to}\:\mathrm{strike}\:\mathrm{the}\:\mathrm{ground}? \\ $$

Answered by mr W last updated on 04/Sep/21

−75=10t−((10)/2)t^2 t^2 −2t−15=0 (t−5)(t+3)=0 ⇒t=5 seconds

$$−\mathrm{75}=\mathrm{10}{t}−\frac{\mathrm{10}}{\mathrm{2}}{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{15}=\mathrm{0} \\ $$$$\left({t}−\mathrm{5}\right)\left({t}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{5}\:{seconds} \\ $$

Commented by nadovic last updated on 05/Sep/21

Thank you Sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

Answered by physicstutes last updated on 04/Sep/21

Using the kinematic equation y = y_0 + v_0 t+(1/2)gt^2 ⇒ y = 75 + 10t−((10)/2)t^2 [assuming g =10 m/s^2 ] when it reaches the ground, y=0 ⇒ 5t^2 −10t−75=0 ⇒ t^2 −2t−15=0 ⇒ (t−5)(t+3)=0 t=5 seconds since t>0

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{kinematic}\:\mathrm{equation} \\ $$$${y}\:=\:{y}_{\mathrm{0}} +\:{v}_{\mathrm{0}} {t}+\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \\ $$$$\Rightarrow\:{y}\:=\:\mathrm{75}\:+\:\mathrm{10}{t}−\frac{\mathrm{10}}{\mathrm{2}}{t}^{\mathrm{2}} \:\:\left[\mathrm{assuming}\:\mathrm{g}\:=\mathrm{10}\:\mathrm{m}/\mathrm{s}^{\mathrm{2}} \right] \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{reaches}\:\mathrm{the}\:\mathrm{ground},\:{y}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{5}{t}^{\mathrm{2}} −\mathrm{10}{t}−\mathrm{75}=\mathrm{0} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow\:\left({t}−\mathrm{5}\right)\left({t}+\mathrm{3}\right)=\mathrm{0} \\ $$$${t}=\mathrm{5}\:\mathrm{seconds}\:\mathrm{since}\:{t}>\mathrm{0} \\ $$

Commented by nadovic last updated on 05/Sep/21

Thank you Sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir} \\ $$

Answered by talminator2856791 last updated on 04/Sep/21

max height = 10×((10)/(9.8)) ÷ 2 + 75 v × (v/(9.8)) ÷ 2 = 10 × ((10)/(9.8)) ÷ 2 + 75 (the point is to find maximum velocity then t follows) v^2 = 1570 v = (√(1570)) t = ((10)/(9.8)) + ((√(1570))/(9.8)) t = 5.06

$$\: \\ $$$$\:\mathrm{max}\:\mathrm{height}\:=\:\mathrm{10}×\frac{\mathrm{10}}{\mathrm{9}.\mathrm{8}}\:\boldsymbol{\div}\:\mathrm{2}\:+\:\mathrm{75} \\ $$$$\:{v}\:×\:\frac{{v}}{\mathrm{9}.\mathrm{8}}\:\boldsymbol{\div}\:\mathrm{2}\:=\:\mathrm{10}\:×\:\frac{\mathrm{10}}{\mathrm{9}.\mathrm{8}}\:\boldsymbol{\div}\:\mathrm{2}\:+\:\mathrm{75} \\ $$$$\:\left(\mathrm{the}\:\mathrm{point}\:\mathrm{is}\:\mathrm{to}\:\mathrm{find}\:\mathrm{maximum}\right. \\ $$$$\left.\:\:\mathrm{velocity}\:\mathrm{then}\:{t}\:\mathrm{follows}\right) \\ $$$$\:\:\:{v}^{\mathrm{2}} \:=\:\mathrm{1570} \\ $$$$\:\:\:{v}\:=\:\sqrt{\mathrm{1570}}\:\: \\ $$$$\:\: \\ $$$$\:{t}\:=\:\frac{\mathrm{10}}{\mathrm{9}.\mathrm{8}}\:+\:\frac{\sqrt{\mathrm{1570}}}{\mathrm{9}.\mathrm{8}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:{t}\:=\:\mathrm{5}.\mathrm{06} \\ $$

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