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Standard XII

Physics

Projectile Time, Height and Range

At what angle...

Question

At what angle with the horizontal should a ball be thrown so that the range $R$ is related to the time of flight $(T)$ as $R = 5 T^{2} ? (Take g = 10 {m/s}^{2})$

30^{\circ}

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45^{\circ}

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60^{\circ}

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90^{\circ}

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Solution

The correct option is B $45^{\circ}$
Given, $\frac{R}{T^{2}} = 5$
We know that
$R = \frac{u^{2} sin 2 θ}{g}; T = \frac{2 u sin θ}{g}$
$∴ \frac{R}{T^{2}} = g \frac{sin 2 θ}{4 {sin}^{2} θ} = \frac{g cot θ}{2} = 5 cot θ$
Hence, $5 = 5 cot θ$
$\Rightarrow θ = 45^{\circ}$

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Similar questions

Q. At what angle with the horizontal should a ball be thrown so that the range

R

is related to the time of flight

(T)

R = 5 T^{2} ? (Take g = 10 {m/s}^{2})

Q. Single Correct Answer Type
At what angle with the horizontal should a ball be thrown so that its range

R

is related to the time of flight as

R = 5 T^{2}

. Take

g = 10 m s^{- 2}

Q. For a projectile projected from ground at an angle

θ

with horizontal,

g T^{2} = \frac{2 R}{\sqrt{3}}

, where

T

is time of fight,

R

is horizontal range of projectile,

g

is acceleration due to gravity. The angle of projection

(θ)

When a particle is projected at an angle to the horizontal, it has range R and time of flight t1. If the same projectile is projected with the same speed at another angle to have the same range and time of flight $t_{2}$ then,

Q. Consider the following statements

A

and

B

and mark the correct option.

A)

The maximum range is

4

times the maximum height attained during its flight in the case of a projectile.

B)

In the case of a projectile the range

R

is related to the time of flight

T

R = 5 T^{2}

. If

g = 10 m s^{- 2}

the angle of projection is

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Projectile Time, Height and Range

Standard XII Physics

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At what angle with the horizontal should a ball be thrown so that the range R is related to the time of flight (T) as R=5T2?(Take g=10 m/s2)

The correct option is B 45∘Given, RT2=5 We know that R=u2 sin 2θg;T=2u sinθg ∴RT2=gsin2θ4 sin2θ=g cotθ2=5 cotθ Hence, 5=5 cotθ ⇒θ=45∘

At what angle with the horizontal should a ball be thrown so that the range $R$ is related to the time of flight $(T)$ as $R = 5 T^{2} ? (Take g = 10 {m/s}^{2})$

The correct option is B $45^{\circ}$
Given, $\frac{R}{T^{2}} = 5$
We know that
$R = \frac{u^{2} sin 2 θ}{g}; T = \frac{2 u sin θ}{g}$
$∴ \frac{R}{T^{2}} = g \frac{sin 2 θ}{4 {sin}^{2} θ} = \frac{g cot θ}{2} = 5 cot θ$
Hence, $5 = 5 cot θ$
$\Rightarrow θ = 45^{\circ}$