Question

For a projectile that lands at the same height as it is released, show that the maximum horizontal distance is reached if the projectile is released at an angle of 45 deg with respect to the horizontal.

Answer 1

Answer:

Explanation:

Given

zero vertical displacement

suppose \theta is the inclination angle and u is the initial velocity

considering horizontal motion

$u_x=u\cos \theta$

range is given by

$R=u\cos \theta \times t$

where t is the time of flight

Now consider vertical motion

$u_y=u\sin \theta$

$y=ut+\frac{1}{2}at^2$

a=acceleration

for zero displacement y=0

$0=u\sin \theta \times t-\frac{1}{2}gt^2$

$t=\frac{2u\sin \theta }{g}$

substitute the value of t in Range

$R=u\cos \theta \times \frac{2u\sin \theta }{g}$

$R=\frac{u^2\sin 2\theta }{g}$

for maximum vale of $R \sin 2\theta$ must be equal to 1

i.e. $\sin 2\theta =1$

$2\theta =\frac{\pi }{2}$

$\theta =\frac{\pi }{4}$