Question

1. A baseball player hits a baseball into the air with an initial vertical velocity of 72 feet per second from a height of 3 feet. ( h = -16t2 +72t +s )

Answer 1

Answer:

You actually do not ask anything here, so i will find the equation for the height and i will use it to find something useful, like the maximum height of the baseball.

For any object in a vertical motion, the only force acting on it is the gravitational force, so the acceleration of the object will be equal to g.

a(t) = -g

where the minus sign is because the gravitational acceleration pulls down.

For the velocity we can integrate over time, and get:

v(t) = -g*t + v0

Where v0 is the initial velocity, in this case v0 = 72 ft/s

v(t) = -g*t + 72ft/s

with this equation we can find the time in wich the height is maximum, we need to find the time where the velocity is 0.

-g*t + 72 = 0

t = 72/g

for the height we can integrate over time again:

h(t) = (-g/2)*t^2 + 72ft/s*t + h0

where h0 is the initial height, in this case h0 = 3ft

here we can also replace the value of g = 32ft/s^2

h(t) = (-16ft/s^2)*t^2 + 72ft/s*t + 3ft

Now we can evaluate this in the time that we previously find, t = 72/g = 72/32 = 2.25s

h(2.25s) = -16*2.25^2 + 72*2.25 + 3 = 85.08 feet