Answer:
Step-by-step explanation:
The formula for projectile motion in this case where we're dealing the "American" measure of gravity in feet per second squared, is
[tex]s(t)=-16t^2+v_{0}+h_{0}[/tex] where s(t) is the height of the object after a certain amount of time has gone by, v₀ is the initial vertical velocity (our unknown in this case), and h₀ is the initial height.
For us, the initial height is 0 because the ball started off on the ground. What we are being asked is what is v₀, when after 2 seconds have gone by and the ball is 94 feet in the air. 94 feet in the air is the position of the ball, s(t) after t seconds go by. What we are being asked to find then, is v₀ when s(2) = 94
[tex]94=-16(2)^2+v_{0}(2)[/tex] which simplifies to
94 = -64 + 2v₀ and
158 = 2v₀ so
v₀ = 79 feet per second
