Can someone please help me

At a time t after being projected from the ground, a projectile is displaced x units horizontally and y units vertically above its point of projection. (Use any variable or symbol stated above along with the following as necessary: g for the acceleration due to gravity.)

What are the horizontal and vertical components of the initial velocity of the projectile?

At the instant the projectile achieves its maximum height above ground level, how far is it displaced horizontally from the launch point?

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adioabiola adioabiola · Answer 1 · 2021-09-15T00:35:50+02:00

The horizontal component of the projectile's initial velocity is Vx = vCosπ

The vertical component of the projectile's initial velocity is Vy = vSinπ

At the instant the projectile achieves its maximum height above ground level, it is displaced X = R/2 = (V²Sin2π)/2g

Assuming the projectile is projected at an angle π with the horizontal...

Therefore, resolving the projectile motion in the horizontal and vertical axis results in:

The horizontal component of the projectile's initial velocity is Vx = vCosπ

The vertical component of the projectile's initial velocity is Vy = vSinπ

Where v is the initial velocity of projection.

The time required to return back to its plane of projection ,T = (2vSinπ)/g

Therefore, the range, R of the projectile,

R = Vy × (2vSinπ)/g

R = vCosπ × (2vSinπ)/g

Recall, 2sinπcosπ = Sin2π

Therefore, R = v²sin2π/g.

Therefore, the horizontal displacement at Maximum height , X = R/2 = (v²Sin2π)/2g.