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

Answer:

a) $t=15 s$

b) $t=7.5s$

c) $f(t)=900ft$

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

$v=\frac{df}{dt}$

$\frac{dt}{dt}=240-32t$

At the highest point its velocity is 0

$0=240-32t$

$t=\frac{240ft/s}{32ft/s^{2} }=7.5s$

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

$f(t)=240(7.5)-16(7.5)^{2}=900ft$