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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 ______.
Options
60 m
71 m
100 m
141 m
Solution
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
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