Question

A ball is thrown at a speed of 40 m/s at an angle of ${60}^0$ with the horizontal. Find

(a) The maximum height reached and

(b) The range of the ball. Take g = 10 $m/s^2.$

Answer 1

Given, u = 40 m/s, a = 9.8 m/s^2, angle of projection, $\theta ={60}^0$

(a) Maximum height,

h = $\frac{u^{2}si{n}^2\theta }{2g}$

= $\frac{{40}^2(sin{60}^0{)}^2}{2\times 10}$ = 60 m

(b) Horizontal range,

X = $\frac{\left(u^{2}sin2\theta \right)}{g}$

= $\frac{{40}^{2}sin2\times {60}^2}{g}=80\sqrt{3}m$