Question

A gun projects a ball at the angle of ${45}^{\circ }$ with the horizontal. If the horizontal range is $39.2$ m, then the ball will rise to

• A. $9.8$ m
• B. $4.9$ m
• C. $2.45$ m
• D. $19.6$ m

Answer 1

The correct option is A $9.8$ m


$v$ is the velocity, $H$ is the maximum height and $R=39.2$ m is the range
$R=\frac{v^2\sin 2\theta }{g}$
$\Rightarrow 39.2=\frac{v^2\sin (2\times 45{)}^{\circ }}{g}\Rightarrow \frac{v^2}{g}=39.2$
$H=\frac{v^2{\sin }^2\theta }{2g}=39.2\times {\sin }^2{45}^{\circ }\times \frac{1}{2}=9.8\text{m}$