Question

a projectile has a range 80m and reaches a maximum height of 20m the angle of elevation at wich the projectile is fired​

Answer 1

Answer:

For the projectile motion:

H/ R = tan θ / 4

H = 20m.

R = 80m.

Tanθ = 4 × 20/80 = 1

θ = Arctan 1

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Answer 2

Given:

Let u be the initial velocity of projection and θ be the angle of the projectile.

Range of the projectile:

$\mathrm{R}=\frac{\mathrm{u}^{2} \sin 2 \theta}{\mathrm{g}} -(1)$

Maximum height of the projectile:

$\mathrm{H}=\frac{\mathrm{u}^{2} \sin ^{2} \theta}{2 \mathrm{~g}}$

Dividing equation (2) by (1)

$\frac{\mathrm{H}}{\mathrm{R}}=\frac{\sin ^{2} \theta}{2 \sin 2 \theta}$

$=\frac{\sin ^{2} \theta}{4 \sin \theta \cos \theta}$

$=\frac{\tan \theta}{4}$

$\frac{20}{80}=\frac{\tan \theta}{4}$

$\frac{1}{4} =\frac{\tan \theta}{4}$

$\theta=\tan ^{-1}\left(\frac{4}{4}\right)\\\theta=\tan ^{-1}(1)$