Question

A cannon ball is fired at a speed of $50m{s}^{-1}$, where the angle of projection is ${30}^{\circ }$. Find the maximum height and also the time of flight of the cannon ball. (Take acceleration due to gravity, $g=10m{s}^{-2}$).

• A. T = 5 s, H = 31.25 m
• B. T = 5 s, H = 45.6 m
• C. T = 11 s, H = 31.25 m
• D. T = 10 s, H = 45.1 m

Answer 1

The correct option is A

T = 5 s, H = 31.25 m

Given , initial velocity, $u=50m{s}^{-1}$ and angle of projecton, $\theta$ = ${30}^{\circ }$.
Also, acceleration due to gravity, $g=10m{s}^{-2}$.
We know, time of flight, T $=\frac{2usin\theta }{g}$
$=\frac{2\times 50\times sin{30}^{\circ }}{10}$
$=2\times$50$\times$$\frac{1}{2}$$\times$$\frac{1}{10}$
$=5s$
Maximum height, H $=\frac{u^{2}si{n}^2\theta }{2g}$
= $\frac{(50{)}^{2}si{n}^{2}30}{2\times 10}$
= 2500 $\times$$\frac{1}{4}$$\times$$\frac{1}{20}$
= $\frac{2500}{4\times 20}$ $=31.25m$