For an object projected from ground with speed u horizontal range is two times the maximum height attained by it. The horizontal range of object is

Question

For an object projected from ground with speed u horizontal range is two times the maximum height attained by it. The horizontal range of object is (A) (2 u2/3 g) (B) (3 u2/4 g) (C) (3 u2/2 g) (D) (4 u2/5 g)

Tardigrade Admin · Accepted Answer

R=24 also, (H/R)=(1/4) tan θ (H/R)=(1/2) ⇒ (1/2)=(1/4) tan θ tan θ=2=(P/B) R=(2 u2 sin θ cos θ/g) R=(2 u2/g)(2/√5) × (1/√5) R=(4 u2/5 g)

Q. For an object projected from ground with speed $u$ horizontal range is two times the maximum height attained by it. The horizontal range of object is

$\frac{2 u ^{2}}{3 g}$

$\frac{3 u ^{2}}{4 g}$

$\frac{3 u ^{2}}{2 g}$

$\frac{4 u ^{2}}{5 g}$

$R = 24$ also, $\frac{H}{R} = \frac{1}{4} tan θ$
$\frac{H}{R} = \frac{1}{2}$
$\Rightarrow \frac{1}{2} = \frac{1}{4} tan θ$
$tan θ = 2 = \frac{P}{B}$
$R = \frac{2 u ^{2} s i n θ c o s θ}{g}$
$R = \frac{2 u ^{2}}{g} \frac{2}{5} \times \frac{1}{5}$
$R = \frac{4 u ^{2}}{5 g}$

Q. For an object projected from ground with speed u horizontal range is two times the maximum height attained by it. The horizontal range of object is

3g2u2​

4g3u2​

2g3u2​

5g4u2​