Question

A ball is projected from the ground at a speed of $ 10\text{ }m{{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}=1.25\,m $
Time to reach maximum height
$ t=\frac{u\sin \theta }{g} $
$ \therefore $ $ t=\frac{10\times \sin 30{}^\circ }{10}=0.5\,s $
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}}=1.25\,m $
$ \therefore $ Height of second ball
$ =1.25+1.25=2.5m $