Question

show that the maximum height reached by a projectile launched at an angle of 45 degrees is one quarter of its range.
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Answer 1

Answer:

Maximum height is given by H = u²sin²θ/2g

And Range is given by R = u²sin2θ/g

Here, u is initial velocity of projectile , θ is angle between initial velocity and horizontal line and g is acceleration due to gravity.

At θ = 45°

Maximum height , H = u²sin²45°/2g = u²(1/√2)²/2g

H = u²/4g -------(1)

Range , R = u²sin2θ/g = u²sin2(45°)/g

R = u²sin90°/g = u²/g ------(2)

dividing (2) ÷ (1)

R/H = 4

⇒ R = 4H

⇒H = R/4

Hence it is clear that maximum height is one quarter of its range at 45

Explanation: