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 ______.

पर्याय

• 60 m

• 71 m

• 100 m

• 141 m

Answer 1

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 100 m.

Explanation:

We know that

Where θ is angle of projection

Given, θ = 15° and R = 50 m

Range, R = `(u^2 sin 2 θ)/g`

Putting all the given values in the formula, we get

⇒ R = `50  m = (u^2 sin(2 xx 15^circ))/g`

⇒ `50 xx g = u^2 sin  30^circ = u^2 xx 1/2`

⇒ `50 xx g xx 2 = u^2`

⇒ `u^2 = 50 xx 9.8 xx 2 = 100 xx 9.8` = 980

⇒ `u = sqrt(980)`

= `sqrt(49 xx 20)`

= `7 xx 2 xx sqrt(5)` m/s

= `14 xx 2.23` m/s

= 31.304 m/s

For θ = 45°, R = `(u^2 sin 2 xx 45^circ)/g = u^2/g` ......(∵ sin 90° = 1) 

⇒ R = `(14sqrt(5))^2/g`

= `(14 xx 14 xx 5)/9.8`

= 100 m