Question

A projectile reaches maximum height of 16 m when projected with angle ${45}^0$ with horizontal.Its horizontal range is:

• A. 4 m
• B. 64 m
• C. 32 m
• D. 8 m

Answer 1

The correct option is A 4 m

Given,

$H=16m$

$\theta ={45}^0$

The maximum height in projectile motion,

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

$16=\frac{u^{2}si{n}^2{45}^0}{2\times 10}$

$u^2=640$

Horizontal range of the projectile,

$R=\frac{u^{2}sin2\theta }{g}$

$R=\frac{640\times sin{90}^0}{10}$

$R=64m$

The correct option is B.