Question

A projectile is thrown with a velocity of $10\sqrt{2}m/s$ at an angle of ${45}^{\circ }$ with horizontal. The interval between the moments when speed is $\sqrt{125}m/s$ is
$(g=10m/s^2)$

• A. $1s$
• B. $1.5s$
• C. $2s$
• D. $0.5s$

Answer 1

The correct option is A $1s$

$u_x=10\sqrt{2}\cos {45}^{\circ }=10m/s{u}_y=10\sqrt{2}\sin {45}^{\circ }=10m/s$
Let $\alpha$ be the angle with the horizontal when speed is $=\sqrt{125}m/s$ ;
As the horizontal component of velocity always remains constant
$10\sqrt{2}\cos {45}^{\circ }=\sqrt{125}\cos \alpha \Rightarrow \cos \alpha =\frac{2}{\sqrt{5}}$
i.e.$v_y=\sqrt{125}\times \sin \alpha =5\sqrt{5}\times \frac{1}{\sqrt{5}}=5m/s$

Using first equation of motion: $v_y=u_y-g\left(t_1\right)\Rightarrow 5=10-10t_1\Rightarrow t_1=\frac{1}{2}s$
During downward journey: $-v_y=u_y-g\left(t_2\right)\Rightarrow -5=10–10t_2\Rightarrow t_2=\frac{3}{2}s$
$\therefore t_2-t_1=1s$
The interval between the moments when speed is $\sqrt{125}m/s$ is $1s$.

Alternative solution

Horizontal component will remain constant
$10\sqrt{2}\cos {45}^{\circ }=\sqrt{125}\cos \alpha \Rightarrow \cos \alpha =\frac{2}{\sqrt{5}}\Rightarrow \sin \alpha =\frac{1}{\sqrt{5}}$
For distance $AB$ in path of projectile,
Time period $T=\frac{2v\sin \alpha }{g}=\frac{2\sqrt{125}\times \frac{1}{\sqrt{5}}}{10}=1s$