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

The nut lands first on gorund

Solution:

The squirrel climbs down the tree in 2 seconds

Given that,

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:

$h(t) = -16t^2 + 8t+24$

$-16t^2 + 8t+24 = 0\\\\-2t^2 + t + 3 = 0\\\\2t^2 - t-3=0\\\\Split\ the\ middle\ term\\\\2t^2 + 2t-3t - 3 = 0\\\\Break\ the\ expression\ into\ groups\\\\(2t^2 + 2t) -(3t+3) = 0\\\\2t(t+1) -3(t+1) = 0\\\\(t+1)(2t-3) = 0\\\\Therefore,\\\\t + 1 = 0\\\\t = -1\\\\2t-3=0\\\\2t = 3\\\\t = \frac{3}{2} = 1.5$

time cannot be negative, so ignore t = -1

Thus the nut takes 1.5 seconds to reach ground

From given,

The squirrel climbs down the tree in 2 seconds

Therefore, the nut lands first on gorund