Question

A ball is projected with a velocity of $\sqrt{17}m/s$ at an angle $\theta$ with the horizontal. Its horizontal range is equal to its maximum height. Velocity of the particle $0.1s$ after it is projected is (in $m/s$)
Take $g=10m/s^2$

• A. $\sqrt{17}$
• B. $\sqrt{20}$
• C. $\sqrt{10}$
• D. $\sqrt{40}$

Answer 1

The correct option is C $\sqrt{10}$

By using the relation $R\tan \theta =4H$
$H\tan \theta =4H(\because R=H$)
$\Rightarrow \tan \theta =4$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{17}}\text{and}\sin \theta =\frac{4}{\sqrt{17}}$
Let $v$ be the velocity of the ball after $0.1s$
$v_x=u_x=u\cos \theta =\sqrt{17}\times \frac{1}{\sqrt{17}}=1m/s$
$v_y=u\sin \theta -gt=\sqrt{17}\times \frac{4}{\sqrt{17}}-10\times 0.1=3m/s$ (negative sign indicates the ball is in its downward journey)
$v=\sqrt{v_x^2+v_y^2}=\sqrt{1^2+3^2}=\sqrt{10}m/s$
Hence, the correct answer is option (c)