Question

A particle is projected from horizontal making an angle ${60}^{\circ }$ with initial velocity $40m{s}^{-1}$. Find the time taken by the particle to make angle ${45}^{\circ }$ from horizontal.

• A. $1.5s$
• B. $2.5s$
• C. $3.5s$
• D. $4.5s$

Answer 1

The correct option is

C

$1.5s$

We know that, at ${45}^{\circ },v_x=v_y$

$v_x=u_x=40\sin {60}^{\circ }=v_y$

Now, $v_y=u_y-gt$

$\therefore t=\frac{u_y-v_y}{g}$

$t=\frac{40(\cos {30}^{\circ }-\sin {60}^{\circ })}{9.8}=1.5s$