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

The squirrel reaches the ground before the nut.

Step-by-step explanation:

To determine whether the squirrel reaches the ground before the nut, we need to compare the time it takes for each to reach the ground. The equation for the height of the squirrel is given as h(t) = -16t^2 + 8t + 24. The squirrel climbs down the tree in 2 seconds, so we can plug in t = 2 into the equation to find the height of the squirrel at that time. h(2) = -64 + 16 + 24 = -24 ft.

Now, we need to find the time it takes for the nut to reach the ground. The height of the nut, h(t), is given by the same equation. We need to find the time when h(t) = 0. Setting the equation equal to zero and solving for t, we get -16t^2 + 8t + 24 = 0. Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), we get t ≈ 2.73 seconds or t ≈ -0.48 seconds. Since time cannot be negative, we disregard the negative solution. Therefore, the nut takes approximately 2.73 seconds to reach the ground.

Comparing the times, we can see that the squirrel reaches the ground before the nut. The squirrel takes 2 seconds to climb down the tree, while the nut takes approximately 2.73 seconds to fall. Therefore, the squirrel reaches the ground before the nut.

Answer 2

Step-by-step 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.