A projectile is projected with initial velocity (6 hati+8 hatj) ms -1 If g=10 m s -2, then horizontal range is

Question

A projectile is projected with initial velocity (6 hati+8 hatj) ms -1 If g=10 m s -2, then horizontal range is (A) 4.8 m (B) 9.6 m (C) 19.2 m (D) 14.0 m

Tardigrade Admin · Accepted Answer

Here, u cos θ=6, u sin θ=8 and g=10 m s -2 ∴ R =(u2 sin 2 θ/g)=(2 u2 sin θ cos θ/g)=(2(u sin θ)(u cos θ)/g) =(2 × 8 × 6/10)=9.6 m

Q. A projectile is projected with initial velocity $(6 \hat{i} + 8 \hat{j}) m s^{- 1}$ If $g = 10 m s^{- 2}$ , then horizontal range is

4.8 m

9.6 m

19.2 m

14.0 m

Here, $u cos θ = 6, u sin θ = 8$ and
$g = 10 m s^{- 2}$
$∴ R = \frac{u ^{2} s i n 2 θ}{g} = \frac{2 u ^{2} s i n θ c o s θ}{g} = \frac{2 ( u s i n θ ) ( u c o s θ )}{g}$
$= \frac{2 \times 8 \times 6}{10} = 9.6 m$

Q. A projectile is projected with initial velocity (6i^+8j^​)ms−1 If g=10ms−2, then horizontal range is