Question

The horizontal 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​

Answer 1

Answer:

100 m

The horizontal range of a projectile fixed at an angle=15°

Range=50m

In dirst case range=50=u2/g(sin2×15°)

50=2g2gu2=100

In 2nd case range=g(sin2×45°)u2=1100(∵sin90°=1)=100m

A ball is projected on an inclined plane, having inclination θ′ ; at an angle ′θ′ with the horizontal. The value of θ for which the range is maximum is,

Two particles are projected with same velocity but at angles of projection 35∘ and 55∘

Answer 2

Answer:

The formula to calculate the range of a projectile is given as:

$\boxed{\bf{ R = \dfrac{u^2\:.\:sin\:2\theta}{g}}}$

According to the question, when the projectile is fired at an angle of 15 degrees, the Range is 50 m. Since the initial velocity is unknown, we can find it using the given information. Substituting the known values we get:

$\implies R = \dfrac{u^2.\:sin(2\times 15)}{10}\\\\\\\implies 50 = \dfrac{u^2.\:sin(30)}{10}\\\\\\\text{Cross multiplying we get:}\\\\\\\implies 50 \times 10 = u^2 \times \dfrac{1}{2}\\\\\\\implies 500 = \dfrac{u^2}{2}\\\\\\\implies 500 \times 2 = u^2\\\\\implies \boxed{ \bf{ u^2 = 1000\:m/s}}$

Now, we are required to find the value of the range, when the angle is 45 degrees. Substituting the values we get:

$\implies R = \dfrac{u^2.\:sin(2\times 45)}{10}\\\\\\\implies R = \dfrac{1000.\:sin(90)}{10}\\\\\\\implies R = \dfrac{1000 \times 1 }{10}\\\\\\\implies R = \dfrac{1000}{10}\\\\\implies \boxed{ \bf{R = 100\:m}}$

Hence the range of the projectile when it is projected at an angle of 45 degrees is 100 m.