A ball is thrown from the location of a horizontal playground with an initial speed at an angle from the -direction. The ball is to be hit by a stone, which is thrown at the same time from the location . The stone is thrown at an angle from the -direc

Question

A ball is thrown from the location $(x_{0}, y_{0}) = (0, 0)$ of a horizontal playground with an initial speed $v_{0}$ at an angle $θ_{0}$ from the $+ x$ -direction. The ball is to be hit by a stone, which is thrown at the same time from the location $(x_{1}, y_{1}) = (l, 0)$ . The stone is thrown at an angle $(180 - θ_{1})$ from the $+ x$ -direction with a suitable initial speed. For a fixed $v_{0}$ , when $(θ_{0}, θ_{1}) = (45 °, 45 °)$ , the stone hits the ball after time $T_{1}$ , and when $(θ_{0}, θ_{1}) = (60 °, 30 °)$ , it hits the ball after time $T_{2}$ . In such a case, ${(\frac{T_{1}}{T_{2}})}^{2}$ is ______ . [2024]

Smart Achievers · Accepted Answer

(2)

For case I: For collision, $\frac{v_{0}}{\sqrt{2}} = \frac{v}{\sqrt{2}}$

$∴$ $v = v_{0}$

$So$ $T_{1} = \frac{L}{\frac{v_{0}}{\sqrt{2}} + \frac{v}{\sqrt{2}}} = \frac{L}{\sqrt{2} v_{0}}$

$For case II: For collision, \frac{v_{0} \sqrt{3}}{2} = \frac{v}{2} \Rightarrow v = \sqrt{3} v_{0}$

$So,$ $T_{2} = \frac{L}{\frac{v_{0}}{2} + v \frac{\sqrt{3}}{2}} = \frac{L}{\frac{v_{0}}{2} + \frac{3 v_{0}}{2}} = \frac{L}{2 v_{0}}$

$So, {(\frac{T_{1}}{T_{2}})}^{2} {(\frac{L / \sqrt{2} v_{0}}{L / 2 v_{0}})}^{2} = {(\sqrt{2})}^{2} = 2$

Solution

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Solution

Ans. (2) For case I: For collision, v02=v2 ∴ v=v0 So T1=Lv02+v2=L2v0 For case II: For collision, v032=v2⇒v=3v0 So, T2=Lv02+v32=Lv02+3v02=L2v0 So, (T1T2)2(L/2v0L/2v0)2=(2)2=2

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