posted by himel .
In a projectile motion, the horizontal range and the maximum height attained by the projectile are equal .
A) what is the launch angle?
B) If everything else slays the same , how should the launch angle, ¦È0, of a projectile be changed for the range of the projectile to be halve?
range = v^2/g sin(2θ)
height = v^2/g sin^2(θ)
so, solve for θ
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thanks a lot sir@Steve