A stone thrown from a point on the ground at $30^{\circ}$ to the horizontal to hit another point on the ground $145 m$ away, fell short by $5 m$ . A second trial at a different angle (but the same speed) again fell short by the same distance. What was the difference in their flight times ?

5 s

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4 s

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3 s

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8 s

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Solution

The correct option is C $3 s$
As the stone hits $5 m$ before the target, the range of projectile is $145 - 5 = 140 m$
In the given question, range being the same, the second throw must have been at $90^{\circ} - 30^{\circ} = 60^{\circ}$ .

$T = \frac{2 u sin θ}{g}$ (time of flight)
$\Rightarrow T_{1} = \frac{2 u sin 30^{\circ}}{g} and T_{2} = \frac{2 u sin 60^{\circ}}{g}$
$\Rightarrow \frac{T_{2}}{T_{1}} = \frac{sin 60^{\circ}}{sin 30^{\circ}} = \sqrt{3}$
$\Rightarrow T_{2} = \sqrt{3} T_{1} \dots \dots (1)$
and
$T_{1} T_{2} = \frac{4 u^{2}}{g^{2}} \times (sin 30^{\circ}) (sin 60^{\circ})$
$= \frac{4 u^{2} (2 sin 30^{\circ} cos 30^{\circ})}{2 g^{2}}$
$= \frac{2 R}{g} = \frac{2 \times 140}{10} = 28 \dots (2)$

Substituting from (1) in (2),
$\Rightarrow \sqrt{3} T_{1}^{2} = 28 \Rightarrow T_{1} = 4.02 \approx 4 s$
$∴ T_{2} \approx 7 s$
$\Rightarrow$ Difference $= 7 - 4 = 3 s$

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A stone thrown from a point on the ground at 30∘ to the horizontal to hit another point on the ground 145 m away, fell short by 5 m. A second trial at a different angle (but the same speed) again fell short by the same distance. What was the difference in their flight times ?

A stone thrown from a point on the ground at $30^{\circ}$ to the horizontal to hit another point on the ground $145 m$ away, fell short by $5 m$ . A second trial at a different angle (but the same speed) again fell short by the same distance. What was the difference in their flight times ?