Question

A body is projected from the ground at an angle of ${30}^o$ with the horizontal with a velocity of $20{ms}^{-1}$. The magnitude of the average velocity of the body during its ascent is: (Acceleration due to gravity $=10{ms}^{-2}$)

• A. $2\sqrt{13}{ms}^{-1}$
• B. $3\sqrt{13}{ms}^{-1}$
• C. $4\sqrt{13}{ms}^{-1}$
• D. $5\sqrt{13}{ms}^{-1}$

Answer 1

Answer: (D)

$5\sqrt{13}{ms}^{-1}$

At the top point, horizontal displacement is equal to half the range, vertical displacement is equal to the maximum height of the projectile and time taken is half the time of flight.

Horizontal displacement is:

$x=\frac{R}{2}=\frac{u^2\sin \left(2\theta \right)/g}{2}$

$x=\frac{{20}^2\sin \left({60}^o\right)}{2\times 10}=10\sqrt{3}m$

Vertical displacement is:

$y=H=\frac{u^2{\sin }^2\theta }{2g}$

$y=\frac{{20}^2{\sin }^2\left({30}^0\right)}{2\times 10}=5m$

Magnitude of total displacement is:

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(10\sqrt{3}{)}^2+5^2}$

$s=5\sqrt{13}m$

Time taken is:

$t=\frac{T}{2}$

$t=\frac{2u\sin \theta /g}{2}$

$t=\frac{20\sin \left({30}^o\right)}{10}=1s$

Average velocity is:

$v=\frac{s}{t}$

$v=5\sqrt{13}m/s$