Question

A shot leaves the gun at the rate of 160m/s calculat the greatest distance to which it could be projected and the heigt to which it would rise

Answer 1

we know from oblique projectile motion,
range is given by $R=\frac{u^2sin2\theta}{g}$
where u is the initial velocity of projectile, $\theta$ is the angle of inclination and g is the acceleration due to gravity.

for maximum value of range ,
$sin2\theta$ should be maximum
we know, maximum value of sine function is 1
so, at $\theta$ = 45° range is maximum.

Therefore, $R_{max}=\frac{u^2sin2(45)}{g}=\frac{u^2}{g}$

here , u = 160m/s and g = 10 m/s²
so, maximum range = (160)²/10 = 2560m

and maximum height , $H=\frac{u^2sin^2\theta}{2g}$

$H=\frac{u^2sin^245^{\circ}}{2g}=\frac{u^2}{4g}$

H = (160)²/40 = 640m

hence, greatest distance = 2560m
and height = 640m