For what angle range becomes double of height in a projectile motion?

1 Answer
Jun 9, 2018

#tan^-1 2#

Explanation:

The range for projectile is #R = u^2/g sin(2 alpha)#
and the maximum height is given by #H = (u^2sin^2 alpha)/(2g)#

Thus, to get #R = 2H# we must have

#u^2/g sin(2 alpha) = 2(u^2sin^2 alpha)/(2g) implies#

#sin(2alpha) = sin^2 alpha implies#

#2sin alpha cos alpha = sin^2 alpha implies#

# tan alpha = 2 #

Thus, the angle of projection must be #tan^-1 2#