Answer to Question #116925 in Classical Mechanics for sridhar

Question

A boy can throw a ball obliquely to a maximum horizontal distance 'X' while standing on the ground.If he throws the same ball from the top of a tower of height ' X ' at an angle of 45° above the horizontal from the foot of the tower the ball hits the ground at a distance (Assume same initial speed)

Answer 1

We have given $X$ is the maximum horizontal range of the given projectile, thus

$$X= v_{0}^2\frac{\sin(2\theta_0)}{g}\implies \theta_0=45^{\circ}$$

where,$v_0 \& \theta_0$ is the initial velocity and angle of the projectile respectively.Hence,

$$X=\frac{v_{0}^2}{g}\implies v_0=\sqrt{Xg} \hspace{1cm}(1)$$

Now, if the same ball projected from height $X$ at the same speed $v_0$ and angle $\theta=45^{\circ}$ ,then time of flight will be, when the ball reaches ground i.e

$$X=v_0\sin(\theta)T-\frac{1}{2}gT^2$$

where, $T$ is the time of flight of the ball.Thus from $(1)$ ,above equation can be simplified to,

$$-X=\sqrt{Xg}\sin(45^{\circ})T-\frac{1}{2}gT^2\\ \implies \sqrt{\frac{Xg}{2}}T-\frac{1}{2}gT^2+X=0\\$$

Clearly, this equation is quadratic in $T$ ,Hence,

$$T=\frac{-\sqrt{\frac{Xg}{2}} \pm \sqrt{\frac{Xg}{2} +4\frac{g}{2}X}}{-\frac{g}{2}}\\ \implies T=\frac{-\sqrt{\frac{Xg}{2}} - \sqrt{5\frac{Xg}{2}}}{-\frac{g}{2}}\\ \implies T=(1+\sqrt{5})\sqrt{\frac{2X}{g}}\hspace{1cm}(\because T\nleq 0)\hspace{1cm}(2)$$

Thus, the range at which given ball will reach at time $T$ is,

$$R=v_0\cos(\theta)T\\ \implies R=\sqrt{\frac{Xg}{2}}\bigg(1+\sqrt{5})\sqrt{\frac{2X}{g}}\bigg) \hspace{1cm}(\because from, (1) \& (2))\\ \implies R=(1+\sqrt{5})X$$

Hence, we are done.