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?

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

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

Dividing through out by 8 we get

Factorizing we get

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.

the answer is -6

step-by-step explanation:

the answer is -6 because its next to the variable m