Two towers $A$ and $B$, each of height $20 \mathrm{~m}$ are situated a distance $200 \mathrm{~m}$ apart. A body thrown horizontally from the top of the tower $A$ with a velocity $20 \mathrm{~ms}^{-1}$ towards the tower $B$ hits the ground at point $P$ and another body thrown horizontally from the…

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Two towers $A$ and $B$, each of height $20 \mathrm{~m}$ are situated a distance $200 \mathrm{~m}$ apart. A body thrown horizontally from the top of the tower $A$ with a velocity $20 \mathrm{~ms}^{-1}$ towards the tower $B$ hits the ground at point $P$ and another body thrown horizontally from the top of tower $B$ with a velocity $30 \mathrm{~ms}^{-1}$ towards the tower $A$ hits the ground at point $Q$. If a car starting from rest from $P$ reaches $Q$ in 10 seconds, then the acceleration of the car is $\left(\right.$ acceleration due to gravity $\left.=10 \mathrm{~ms}^{-2}\right)$

Aicharya · Accepted Answer

Given, height of both towers is same, $h_1=h_2=h$.

Time of flight, $t=\sqrt{\frac{2 h}{g}}$ will be same $ t=\sqrt{\frac{2 \times 20}{10}}=\sqrt{4}=2 \mathrm{~s} $ $\Rightarrow$ Displacement in horizontal direction from tower $A$ to point $P=u_A t$ $ =20 \times 2=40 \mathrm{~m} $ $\Rightarrow$ Displacement in horizontal direction from tower $B$ to point $Q=u_B t$ $ =30 \times 2=60 \mathrm{~m} $ So, distance between point $P$ and $Q$ $ \begin{aligned} & =200-(40+60) \\ & =100 \mathrm{~m} \end{aligned} $ Given, distance between $P$ and $Q$ is covered by car in $10 \mathrm{~s}$, so using $ \begin{aligned} s= & u t+\frac{1}{2} a t^2 \\ 100= & 0 \times 10+\frac{1}{2} a(10)^2 \\ & {[\because u=0, \text { as car starts from rest }] } \\ a= & 2 \end{aligned} $ Acceleration, $a=2 \mathrm{~m} / \mathrm{s}^2$

Two towers $A$ and $B$, each of height $20 \mathrm{~m}$ are situated a distance $200 \mathrm{~m}$ apart. A…

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