Question

Show that the maximum height reached by a projectile launched at an angle of 45 degree is one quarter of its range.

Answer 1

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°

Answer 2

Answer:

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