A particle is projected from a horizontal plane with a velocity of 8 √2 m s -1 at an angle θ. At highest point its velocity is found to be 8 m s -1. Its range will be ( g =10 m s -2)

Question

A particle is projected from a horizontal plane with a velocity of 8 √2 m s -1 at an angle θ. At highest point its velocity is found to be 8 m s -1. Its range will be ( g =10 m s -2) (A) 3.2 m (B) 4.6 m (C) 6.4 m (D) 12.8 m

Tardigrade Admin · Accepted Answer

At highest point, velocity of projectile, v=u cos θ ∴ 8=8 √2 cos θ or cos θ=(1/√2) or θ=45° Range =(u2 sin 2 θ/g) =((8 √2)2 sin 2 × 45°/10) =(64 × 2/10)=12.8 m

Q. A particle is projected from a horizontal plane with a velocity of $82 m s^{- 1}$ at an angle $θ$ . At highest point its velocity is found to be $8 m s^{- 1}$ . Its range will be $(g = 10 m s^{- 2})$

3.2 m

4.6 m

6.4 m

12.8 m

At highest point, velocity of projectile, $v = u cos θ$
$∴ 8 = 82 cos θ$
or $cos θ = \frac{1}{2}$ or $θ = 4 5^{\circ}$
Range $= \frac{u ^{2} s i n 2 θ}{g}$
$= \frac{( 8 2 ) ^{2} s i n 2 \times 4 5 ^{\circ}}{10}$
$= \frac{64 \times 2}{10} = 12.8 m$

Q. A particle is projected from a horizontal plane with a velocity of 82​ms−1 at an angle θ. At highest point its velocity is found to be 8ms−1. Its range will be (g=10ms−2)