Question

When the angle of projection is $75^{\circ }$, a ball falls 10 m short of the target. When the angle of projection is $45^{\circ }$, it falls 10 m ahead of the target. Both are projected from the same point with the same speed in the same direction, the distance of the target from the point of projection is

• A. 15 m
• B. 30 m
• C. 45 m
• D. 10 m

Answer 1

Answer: (B)

Let $d$ be the distance of the target from the point of projection.
$\therefore \frac{u^2\sin \left(2\times 75^{\circ }\right)}{g}=d-10$
or $\frac{u^2}{2g}=d-10\dots (i)$
and $\frac{u^2\sin \left(2\times 45^{\circ }\right)}{g}=d+10$
or $\frac{u^2}{g}=d+10\dots (ii)$
Divide (i) by (ii), we get
$\frac{d-10}{d+10}=\frac{1}{2}$ or
$d=30m$