A small ball is launched directly upward from ground level with an initial velocity of 12 m/s. What is the ball's maximum height above the ground? (Assume no air resistance.)
+561 Use the formula:
\(v^2=u^2+2as\)
Where v is final velocity, u is intitial velocity, a is acceleration, and s is distance.
v = 0 because the ball will stop moving upwards as it reaches it's peak height.
u = 12m/s as given in the question.
a = -9.81m/s/s because gravity is the force acting against the ball.
s is what we want to find.
\(0^2=12^2+2\times-9.81\times s\)
\(s=7.3m\)
