Question

A squirrel is 24 feet up in a tree and tosses a nut out of the tree with an initial velocity of 8 feet per second. The nuts height,h, at time t seconds can be represented by the equation h(t)=-16t^2 + 8t +24. If the squirrel climbs down the tree in 2 seconds, does it reach the ground before the nut?

Answer 1

Answer:

Therefore,

Yes , Nut reaches faster than squirrel  by half second.

Step-by-step explanation:

The nuts height,h, at time t seconds can be represented by the equation $h(t)=-16t^{2} + 8t +24$

To Find:

If the squirrel climbs down the tree in 2 seconds, does it reach the ground before the nut = ?

Solution:

For nut to reach the ground, Put h(t) = 0

$0=-16t^{2} + 8t +24$

Dividing through out by 8 we get

$2t^{2} - t -3=0$

Factorizing we get

$2t^{2}-3t+2t-3=0\\\\t(2t-3)+1(2t-3)=0\\\\(t+1)(2t-3)=0\\\\t = -1\ \ or\ \ t=\dfrac{3}{2}=1.5$

As time cannot be negative

t = 1.5 seconds nut reaches at ground.

The squirrel climbs down the tree in 2 seconds .....Given

Difference of time = 2 - 1.5 = 0.5 second

Therefore,

Yes , Nut reaches faster than squirrel  by half second.

Answer 2

Answer: yes squirrel reaches the ground