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?

1 Answer

Answer:

#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#