Question

A ball is projected from the ground at a speed of $10m{s}^{-1}$ making an angle of $30{}^{\circ }$ with the horizontal. Another ball is simultaneously released from a point on the vertical line along the maximum height of the projectile. The initial height of the second ball is $(g=10m{s}^{-2})$

• A. 6.25m
• B. 2.5m
• C. 3.75m
• D. 5m
• E. 1.25m

Answer 1

Answer: (B)

Maximum height of projectile
$h=\frac{u^2{\sin }^2\theta }{2g}$
$\therefore$ $h=\frac{{(10)}^2\times {\sin }^2(30{}^{\circ })}{2\times 10}$
$=\frac{5}{4}=1.25m$
Time to reach maximum height
$t=\frac{u\sin \theta }{g}$
$\therefore$ $t=\frac{10\times \sin 30{}^{\circ }}{10}=0.5s$
So, distance of vertical fall in 0.5 s
$s=\frac{1}{2}g{t}^2$ or $s=\frac{1}{2}\times 10\times {(0.5)}^2=1.25m$
$\therefore$ Height of second ball
$=1.25+1.25=2.5m$