Posted by nam maurer on .
A projectile was launched 64° above the horizontal, attaining a height of 10 m. What is the projectile's initial speed?
(Average vertical velocity) * (time of flight to max height) = Maximum height H
(Vo*sinA/2)*(VosinA/g) = H
Vo^2*sin^2A/(2g) = H
Solve for Vo