Solveeit Logo

Question

Physics Question on Motion in a plane

From the ground, a projectile is fired at an angle of 6060 degrees to the horizontal with a speed of 20m/s120\, m/s^{-1}. Take acceleration due to gravity as 10m/s210 \, m/s^{-2}. The horizontal range of the projectile is

A

103m10 \sqrt{3}\, m

B

20 m

C

203m20 \sqrt{3}\, m

D

403m40 \sqrt{3}\, m

Answer

203m20 \sqrt{3}\, m

Explanation

Solution

The horizontal range
R=u2sin2θgR=\frac{u^{2} \sin 2 \theta}{g}
Given, u=20m/su=20 \,m / s
θ=60\theta=60^{\circ}
g=10m/s2g=10 \,m / s ^{2}
R=(20)2sin(2×60)10R=\frac{(20)^{2} \sin \left(2 \times 60^{\circ}\right)}{10}
=20×2010×32=203m=\frac{20 \times 20}{10} \times \frac{\sqrt{3}}{2}=20 \sqrt{3} \,m