Question

A projectile is shot from the ground at an angle of `pi/6` and a speed of `4 m/s`. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Answer 1

The maximum height is `=0.2m`

The horizontal distance is `=0.69m`

Explanation:

The maximum height is when the vertical component of the velocity `=0`

We use the equation

`v^2=u^2-2gy`

`0=(v_0sintheta)^2-2gy`

`y=(4*sin(pi/6))^2/(2g)`

`y_(max)=2/g=2/10=0.2m`

The time to reach the greatest height is `t=v_0sintheta/g`

`t=4*sin (pi/6)/10=0.2s`

The horizontal distance is `x=v_0costheta*t`

`=4*cos(pi/6)*0.2=0.8*sqrt3/2=0.4sqrt3=0.69m`

graph{y=(-5x^2/12)+0.58x [-0.819, 1.881, -0.032, 1.318]}