Question

A projectile is thrown with velocity $10\sqrt{2}m/s$ at angle $45$ to horizontal. Calculate time interval between the moments when velocity is $12.5m/s$. $($$g=10m/s^2$$)$ -

• A. $1.0s$
• B. $1.5s$
• C. $2.0s$
• D. $0.5s$

Answer 1

Answer: (B)

$1.5s$

Given : $u=10\sqrt{2}m/s$

$\therefore$ $u_x=ucos{45}^o=10m/s$ and $u_y=usin{45}^o=10m/s$

Let the velocities of the projectile be $V_x$ and $V_y$ in vertical and horizontal directions respectively.

$\therefore$ $V_x=u_x=10m/s$

Using $V^2=V_x^2+V_y^2$

$\therefore$ ${12.5}^2={10}^2+V_y^2$ $⟹V_y=\pm 7.5m/s$

y direction : $V_y=u_y-gt$ where $g=10m/s^2$

Upward motion : $7.5=10-\left(10\right)t_1$ $⟹t_1=0.25$ s

Downward motion : $-7.5=10-\left(10\right)t_2$ $⟹t_2=1.75$ s

$⟹$ Time interval $\Delta t=t_2-t_1=1.75-0.25=1.5$ s