Question

A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and falls at a distance of 14 m from the wall. Find the magnitude of the initial velocity (in m/s) of the ball.

• A. $\sqrt{682}$
• B. $\sqrt{382}$
• C. $\sqrt{182}$
• D. $\sqrt{82}$

Answer 1

The correct option is C $\sqrt{182}$

The ball passes through the point P(4, 4). Also rang = 4 + 14 = 18 m.
The trajectory of the ball is,
$y=xtan\theta (1-\frac{x}{R})$
Now x = 4m, y = 4m and R = 18 m
$\therefore 4=4tan\theta [1-\frac{4}{18}]=4tan\theta \frac{7}{9}$
or $tan\theta =\frac{9}{7}$ $\Rightarrow$ $\theta =ta{n}^{-1}\frac{9}{7}$ And $R=\frac{2u^{2}sin\theta cos\theta }{g}$
or $18=\frac{2}{9.8}\times u^2\times \frac{9}{\sqrt{130}}\times \frac{7}{\sqrt{130}}\Rightarrow u=\sqrt{182}$