Question

2. You throw a softball straight up into the air at a velocity of 30 feet per second. You release the softball at a height of 5.8 feet and catch it when it falls back to a height of 6.2 feet. Use the position equation to write a mathematical model for the height of the softball

Answer 1

h(t)=-16t^2+30t+5.8

Step-by-step explanation

You first need to use the equation: h(t)=-16t^2+vt+s

V= the initial velocity in feet/second

T= time in motion

H=height at a given time

S= initial height

Therefore,

V= 30ft (per second)

S=5.8

Which is

h(t)=-16t^2+30t+5.8

Answer 2

`-16t^2+30t+5.8`

Explanation:

We are given that

Initial velocity=30 ft/s

Initial position=
`s_0=5.8 ft`

We have to write a mathematical model for height of the softball by using the position equation

The position equation on Earth's surface is given by

`h(t)=-16t^2+v_0t+s_0`

Where
`v_0=` Initial velocity

`s_0`=Initial position

Substitute the values in the equation then, we get

`h(t)=-16t^2+30t+5.8`

Hence, the mathematical model for height of the softball is given by

`-16t^2+30t+5.8`