Question

20 POINTS

Answer 1

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

$s(t)=-16t^2+v_{0}+h_{0}$  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

$94=-16(2)^2+v_{0}(2)$ which simplifies to

94 = -64 + 2v₀ and

158 = 2v₀ so

v₀ = 79 feet per second

Answer 2

Answer:

79 feet/sec

Step-by-step explanation:

s = ut + ½at²

a = -32

s = 94

t = 2

94 = 2u -16(2)²

2u = 158

u = 79