A ball is projected from the ground with a speed 15 ms-1 at an angle θ with horizontal so that its range and maximum height are equal. Then ‘tanθ’ will be equal to

H=u2sin2θ2g and R=u2sin2θgAccording to questionR=H⇒u2sin2θ2g=2u2snθcosθg⇒tanθ=4

A ball is projected from the ground with a speed 15 ms-1 at an angle θ with horizontal so that its range and maximum height are equal. Then ‘tanθ’ will be equal to

Ask & Answer Question

physics ncert exemplar solutions class 12th chapter two Projectile Motion Physics Motion in Plane Physics Class 11th

A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then ‘tanθ’ will be equal to

Option 1 -

$\frac{1}{4}$

Option 2 -

$\frac{1}{2}$

Option 3 -

Option 4 -

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

0 Upvotes 0 Downvotes

Similar Questions for you

A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

alok kumar singh

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$= m v (\frac{\sqrt[]{3}}{2}) * \frac{v^{2} s i n^{2} 30}{2 g}$

$= \sqrt[]{3} \frac{m v^{3}}{16 g}$

The trajectory of a projectile in an vertical plane is y = ax - bx², where a and b are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection q and the maximum height attained H are respectively given by:

alok kumar singh

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α \times \frac{α}{2 β} - β \times {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ - $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

A fighter jet is flying horizontally at a certain with a speed of 200ms^-1. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle (θ)with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of (θ)will be _____________.

Raj Pandey

B → fighter jet

A → anti-Air craft gun