Jackie the squirrel is on the roof of morrow library. she throws a peanut nearly vertically upward from a point near the edge of the roof. it just misses the edge of the roof on the way down (when it is in free fall) and passes a point a distance 10.0 m below its starting point at a time 4.75 s after it leaves jackie's paw. air resistance may be ignored. use the magnitude of the acceleration due to gravity as 9.81m/s and report answers to 3 sig figs for parts a-e a) what is the initial velocity of the peanut? b) how high does it rise above its starting point? c) what is the magnitude of its velocity at the highest point? d) what is the magnitude of its acceleration at the highest point? e) what is the direction of its acceleration at the highest point?

A) The initial velocity of the peanut is 21.2 m/s

B) The highest point of the peanut is 22.9 m above the throwing point.

C) The magnitude of the velocity at the highest point is 0.

D)  In this case, the magnitude of the acceleration is always 9.81 m/s².

E) In this case, the direction of the acceleration is always downwards perpendicular to the ground.

Explanation:

The height and velocity of the peanut can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

A) We know that after 4.75 s the height of the peanut is 10.0 m below the throwing point. Then, placing the origin of the frame of reference at the throwing point, we can write the equation of height as follows:

y = v0 · t + 1/2 · g · t²  

Let´s replace with the data and solve for v0:

-10 m = v0 · 4.75 s - 1/2 · 9.81 m/s² · (4.75 s)²

(-10 m + 1/2 · 9.81 m/s² · (4.75 s)²) / 4.75 s = v0

v0 = 21.2 m/s

The initial velocity of the peanut is 21.2 m/s

B) At the highest point, the velocity of the peanut is 0, so using the equation of velocity we can find the time at which the peanut is at its highest point:

v = v0 + g · t

0 = 21.2 m/s - 9.81 m/s² · t

-21.2 m/s / - 9.81 m/s² = t

t = 2.16 s

Now we can calculate the heigt at time t = 2.26 s:

y = y0 + v0 · t + 1/2 · g · t²

y = 0m + 21.2 m/s · 2.16 s - 1/2 · 9.81 m/s² · (2.16)²

y = 22.9 m

The highest point of the peanut is 22.9 m above the throwing point.

C) The magnitude of the velocity at the highest point is 0 because, at that instant, the peanut does not go up nor down.

D) The magnitude of the acceleration is always 9.81 m/s² because the only force acting over the peanut is the gravity (ignoring air resistance).

E) The direction of the acceleration is always downwards perpendicular to the ground. If we consider upwards as positive, then the acceleration of gravity will be -9.81 m/s².

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