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Q. 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

Motion in a Plane

Solution:

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 $\frac{u^{2}}{2 g}=d-10 \,\,\,\,\,\dots(i)$
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
$d=30 \,m$