Asked by nam maurer

A projectile was launched 64° above the horizontal, attaining a height of 10 m. What is the projectile's initial speed?

Answers

Answered by drwls
(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
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