n projectile motion if range is the maximum, then what will be the height?.

Answer 1

recall that maximum range occurs at angle = 45 degrees.
also, the max height in projectile motion is given by:
y,max = (Vo)^2 * sin^2 (theta)/(2g)
where
Vo = initial velocity
theta = angle of release
g = acceleration due to gravity = 9.8 m/s^2

note that max height occurs at theta = 90 because sin^2 (theta) = 1 ---(when the object is thrown upwards, and thus the range is zero):
y,max = (Vo)^2 / 2g

but at theta=45,
y = (Vo)^2 * sin^2 (45) / 2g
y = (Vo)^2 * (sqrt(2)/2)^2 / 2g
y = 1/2 * [(Vo)^2 / 2g]
y = 1/2 * y,max

at 45 degree angle release, the height is only half of the maximum it can reach (which is at 90 degree angle release)

hope this helps~ :)