Question

A stone thrown from a point on the ground at ${30}^{\circ }$ to the horizontal to hit another point on the ground $145\text{m}$ away, fell short by $5\text{m}$. 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. $5\text{s}$
• B. $4\text{s}$
• C. $3\text{s}$
• D. $8\text{s}$

Answer 1

The correct option is C $3\text{s}$

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

$T=\frac{2u\sin \theta }{g}$ (time of flight)
$\Rightarrow T_1=\frac{2u\sin {30}^{\circ }}{g}\text{and}{T}_2=\frac{2u\sin {60}^{\circ }}{g}$
$\Rightarrow \frac{T_2}{T_1}=\frac{\sin {60}^{\circ }}{\sin {30}^{\circ }}=\sqrt{3}$
$\Rightarrow T_2=\sqrt{3}{T}_1\dots \dots \left(1\right)$
and
$T_1{T}_2=\frac{4u^2}{g^2}\times \left(\sin {30}^{\circ }\right)\left(\sin {60}^{\circ }\right)$
$=\frac{4u^2\left(2\sin {30}^{\circ }\cos {30}^{\circ }\right)}{2g^2}$
$=\frac{2R}{g}=\frac{2\times 140}{10}=28\dots \left(2\right)$

Substituting from (1) in (2),
$\Rightarrow \sqrt{3}{T}_1^2=28\Rightarrow T_1=4.02\approx 4\text{s}$
$\therefore T_2\approx 7\text{s}$
$\Rightarrow$ Difference $=7-4=3\text{s}$