Two projectile thrown at 30° and 45° with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

Question

Two projectile thrown at 30° and 45° with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is (A) 1: √2 (B) 2: 1 (C) √2: 1 (D) 1: 2

Tardigrade Admin · Accepted Answer

Time taken to reach maximum height t =( u sin θ/ g ) ∴ ( u 1 sin θ1/ g )=( u 2 sin θ2/ g ) ⇒ u 1 sin 30= u 2 sin 45 ⇒ ( u 1/ u 2)=(1 / √2/1 / 2)=(√2/1)

Q. Two projectile thrown at $3 0^{\circ}$ and $4 5^{\circ}$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

$1 : 2$

$2 : 1$

$2 : 1$

$1 : 2$

Time taken to reach maximum height
$t = \frac{u s i n θ}{g}$
$∴ \frac{u _{1} s i n θ _{1}}{g} = \frac{u _{2} s i n θ _{2}}{g}$
$\Rightarrow u_{1} sin 30 = u_{2} sin 45$
$\Rightarrow \frac{u _{1}}{u _{2}} = \frac{1/ 2}{1/2} = \frac{2}{1}$

Q. Two projectile thrown at 30∘ and 45∘ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

1:2​

2:1

2​:1

1:2

Time taken to reach maximum height t=gusinθ​ ∴gu1​sinθ1​​=gu2​sinθ2​​ ⇒u1​sin30=u2​sin45 ⇒u2​u1​​=1/21/2​​=12​​

Q. Two projectile thrown at $3 0^{\circ}$ and $4 5^{\circ}$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

$1 : 2$

$2 : 1$

$2 : 1$

$1 : 2$

Time taken to reach maximum height
$t = \frac{u s i n θ}{g}$
$∴ \frac{u _{1} s i n θ _{1}}{g} = \frac{u _{2} s i n θ _{2}}{g}$
$\Rightarrow u_{1} sin 30 = u_{2} sin 45$
$\Rightarrow \frac{u _{1}}{u _{2}} = \frac{1/ 2}{1/2} = \frac{2}{1}$