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

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Answer 1 · 2018-06-09T02:32:27

#tan^-1 2#

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#