Question

A squirrel is 24 ft up in a tree and tosses a nut out of the tree with an initial velocity of 8 ft per second. The nuts height, h, at time t seconds can be represented by the equation h(t)=-16t2+8t+24. If the squirrel climbs down the tree in 2 sec, does it reach the ground before the nut?

Answer 1

Answer:

Explanation:

when t=2

$-16t^{2} +8t+24=0 gives\\-2t^2+t+3=0\\-2t^2+3t-2t+3=0\\t(-2t+3)+1(-2t+3)=0\\(-2t+3)(t+1)=0\\t=-1,\frac{3}{2} \\as the squirrel takes 2 sec so it reaches late on the ground.$

Answer 2

Explanation:

It is given that,

Position of the squirrel, $y_o=24\ ft$

Initial speed of the squirrel, u = 8 ft/s

To find,

If the squirrel climbs down the tree in 2 sec, does it reach the ground before the nut

The position of squirrel as a function of time t is given by :

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

Position at t = 2 seconds will be :

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

h(2) = -24 m

At t = 2 s, the nut will hit the ground first than the squirrel.