Question

The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45° its range will be
(a) 25 m
(b) 37 m
(c) 50 m
(d) 100 m Concept of Physics - 1 , HC VERMA , Chapter "Rest and Motion : Kinematics

Answer 1

Correct option d: 100m
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Explanation :
In FIRST CASE range =50 =u²/g(sin2 x 15°)
=u²/2g
u²/g=100
In second case range =u²/g (sin 2 x 45°)
 =[u²/g ]sin90=100m

Answer 2

Answer:

Range, R = 100 m

Explanation:

It is given that,

Range of projectile, R₁ = 50 m

Angle 1, $\theta_1=15^0$

Angle 2, $\theta_2=45^0$

We know that the range of the projectile is given by :

$R_1=\dfrac{v^2sin\ 2\theta_1}{g}$............(1)

$R_2=\dfrac{v^2sin\ 2\theta_2}{g}$.............(2)

Diving equation 1 and 2 we get,

$\dfrac{R_1}{R_2}=\dfrac{sin\ 2\theta_1}{sin\ 2\theta_2}$

$\dfrac{50}{R_2}=\dfrac{sin\ 2(15)}{sin\ 2(45)}$

$R_2=50\times \dfrac{sin\ 2(45)}{sin\ 2(15)}$

R₂ = 100 m

Hence, the correct option is (d) " 100 m ".