Question

If a projectile is shot at an angle of `(5pi)/12` and at a velocity of `1 m/s`, when will it reach its maximum height?

Answer 1

`0.04755416421469008\ m`

Explanation:

The maximum height `H_{max}` obtained by a projectile shot at an angle `\theta={5\pi}/12` with initial velocity `u=1\ m/s` is given as follows

`v^2=u^2+2as` (from II equation of motion)

`0^2=(u\sin\theta)^2-2gH_{max}`

`H_{max}=\frac{u^2\sin\theta}{2g}`

`H_{max}=\frac{1^2\sin^2({5\pi}/12)}{2\times 9.81}`

`=0.04755416421469008\ m`