Abaseball is popped straight up into the site and has a hang time of 6.25s determine the height to which the ball rises before it reaches its peak.

47.9 m

Explanation:

Given:

t = 6.25/2 = 3.125 s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

Δy = vt − ½ at²

Δy = (0 m/s) (3.125 s) − ½ (-9.8 m/s²) (3.125 s)²

Δy = 47.9 m

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer