Question

A body is thrown with a velocity of $10m/s$ at an angle of ${45}^o$ to the horizontal. The radius of curvature of its trajectory in $t=1/\sqrt{2}$sec after the body began to move is:

• A. $0m$
• B. $2.5m$
• C. $5m$
• D. None

Answer 1

The correct option is B $2.5m$

We know maximum height that a projectile can attain $H=\frac{v_0^2{\sin }^2\theta }{2g}$=radius of curvature of trajectory.

Given $v_0=10$ m/s and $\theta ={45}^0$and $g=10m{s}^2$

Thus $H=\frac{100\times 1}{4\times 10}=2.5$ m