# Find the angle of projection at which horizontal range and maximum height are equal

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by TetaBanik917

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by TetaBanik917

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If you remember the following formula : H / R = tan Ф / 4 , then its quick.

* H = R => Ф = tan⁻¹ 4.*

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__Derivation:__

In two dimensional projection of a particle in Earth's gravitational field, we have the following equations :

x = u Cos Ф * t --- (3)

y = u Sin Ф * t - 1/2 g t² --- (1)

v_x = velocity in horizontal direction : u * cosФ

v_y = vertical velocity upwards = u SinФ - g t

v_y = 0 at t = u SinФ/ g

Then by (1) above, y at t = u SinФ/ g is equal to = H

* H = u² Sin²Ф /2g ---(2) *

In the time duration t = u SinФ/g, the projectile travels, a distance of half the Range. Using (3) we get :

R/2 = u² CosФ SinФ / g

* R = u² Sin2Ф / g -- (4)*

From (2) and (4) we get :

* H / R = Tan Ф / 4 ---- (5)*

If H = R are equal, then the angle of projection = tan⁻¹ 4 = 75.96 deg.

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sin A = 4 cos A

Sin² A = 16 Cos² A

cos² A = 1/17

Cos A = 1/√17

Sin A = 4/√17

===============

In two dimensional projection of a particle in Earth's gravitational field, we have the following equations :

x = u Cos Ф * t --- (3)

y = u Sin Ф * t - 1/2 g t² --- (1)

v_x = velocity in horizontal direction : u * cosФ

v_y = vertical velocity upwards = u SinФ - g t

v_y = 0 at t = u SinФ/ g

Then by (1) above, y at t = u SinФ/ g is equal to = H

In the time duration t = u SinФ/g, the projectile travels, a distance of half the Range. Using (3) we get :

R/2 = u² CosФ SinФ / g

From (2) and (4) we get :

If H = R are equal, then the angle of projection = tan⁻¹ 4 = 75.96 deg.

======

sin A = 4 cos A

Sin² A = 16 Cos² A

cos² A = 1/17

Cos A = 1/√17

Sin A = 4/√17