Question

A pomegranate is thrown from ground level straight up into the air at time t=0 with velocity 240 feet per second. Its height in feet at t seconds is f(t)=-16 t^2+240 t. Find the time it hits the ground and the time it reaches its highest point. Hits the ground when t= 4 second (include units) Reaches highest point when t= 2 second (include units) Maximum height =

Answer 1

a)
`t=15 s`

b)
`t=7.5s`

c)
`f(t)=900ft`

Step-by-step explanation:

From the exercise we have the equation of height and the initial velocity of the pomegranate

`f(t)=240t-16t^(2)`

a)To find the time that it hits the ground we know that its position is 0

`0=240t-16t^(2)`

Solving for t using quadratic equation

`t=\frac{-b±\sqrt{b^(2)-4ac } }{2a}`

`a=-16\\b=240\\c=0`

`t=0s` or
`t=15s`

Since time can not be 0, the answer is t=15s

b) To find the time that it reaches its highest point we need to analyze the velocity at that time

`v=(df)/(dt)`

`(dt)/(dt)=240-32t`

At the highest point its velocity is 0

`0=240-32t`

`t=(240ft/s)/(32ft/s^(2) )=7.5s`

c) To find the maximum height we need to use f(t) with t=7.5

`f(t)=240(7.5)-16(7.5)^(2)=900ft`