The range of a projectile at an angle $θ$ is equal to half of the maximum range if thrown at the same speed. The angle of projection $θ$ is given by :

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Insufficient data

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$R = \frac{u^{2} s i n 2 θ}{g} = \frac{u^{2}}{2 g}$
$\Rightarrow s i n 2 θ = \frac{1}{2}$
$\Rightarrow 2 θ = 30^{\circ}$
$\Rightarrow θ = \frac{30^{\circ}}{2} = 15^{\circ}$

Suggest Corrections

Similar questions

Q. The range of a projectile at an angle

θ

is equal to half of the maximum range if thrown at the same speed. The angle of projection is given by

Q. The maximum height attained by a projectile when thrown at an angle

θ

with the horizontal is found to be half the horizontal range. Then,

θ

is equal to:

for a projectile thrown up an incline plane (making an angle $θ$ with
horiontal), maximum range on the incline
can be achieved by projecting at an angle .. from the horizontal ( $θ$ is acute).

Q. A projectile thrown with an initial speed

u

and the angle of projection

15^{o}

to the horizontal has a range

R

. If the same projectile is thrown at an angle of

45^{o}

to the horizontal with speed

2 u

, what will be its range?

Q. In case of a projectile if the maximum vertical height H is equal to horizontal range R then angle of projection

θ

Join BYJU'S Learning Program

The range of a projectile at an angle θ is equal to half of the maximum range if thrown at the same speed. The angle of projection θ is given by :

The correct option is A R=u2sin2θg=u22g ⇒sin2θ=12 ⇒2θ=30∘ ⇒θ=30∘2=15∘

The range of a projectile at an angle $θ$ is equal to half of the maximum range if thrown at the same speed. The angle of projection $θ$ is given by :

The correct option is A
$R = \frac{u^{2} s i n 2 θ}{g} = \frac{u^{2}}{2 g}$
$\Rightarrow s i n 2 θ = \frac{1}{2}$
$\Rightarrow 2 θ = 30^{\circ}$
$\Rightarrow θ = \frac{30^{\circ}}{2} = 15^{\circ}$