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  1. a skydiver jumps from a plane and falls freely for 12 s before opening her parachute. what is their instantaneous speed at time 12 sec?
  2. an apple falls from rest from a tree for 1.5 s.

a. what is its acceleration? explain.
b. what is its final speed at the end of 1.5 s?
c. how far has it fallen?

  1. a rock is dropped off a cliff and falls for 7 s before it hits the ground.

a. what is its acceleration of fall?
b. what is its speed at the end of 7 s?
c. how far does it fall at the end of 7 s?

  1. a rock is thrown off a cliff with an initial downward velocity of 15 m/s and then falls for 5 s before it hits the ground.

a. what is its acceleration?
b. what is its final speed at the end of 5 s?
c. what distance has it traveled at the end of 5 s?

  1. a baseball is thrown into the air with an initial upward velocity of 35 m/s.

a. what is its acceleration?
b. what is the velocity at its highest point?
c. what is the velocity of the ball as it returns to the point at which it was thrown?

  1. a toy rocket is shot straight up into the air with an initial speed of 45.0 m/s.

c. how long does it take for the rocket to reach its highest point?
d. how high does the rocket rise above the ground?

  1. a squirrel drops an acorn from a tree branch 8.00 m from the ground.

a. how long does it take to hit the ground?
b. what is its velocity as it hits the ground?

  1. labron jumps to dunk a basketball with an initial velocity of 5 m/s.

a. how high does he manage to jump?
b. at maximum height, what is his velocity?
c. what is his hang time?

Explanation:

Step1: Identify the relevant formula for free - fall motion

The formula for the instantaneous speed of an object in free - fall is $v = v_0+at$, where $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time. In free - fall near the surface of the Earth, $a = g= 9.8\ m/s^2$ (assuming no air - resistance) and when the object starts from rest, $v_0 = 0$.

Step2: Solve problem 1

For the skydiver, $v_0 = 0$ (starts from rest), $a = g = 9.8\ m/s^2$, and $t = 12\ s$. Using the formula $v=v_0 + at$, we substitute the values: $v=0+9.8\times12$.
$v = 117.6\ m/s$

Step3: Solve problem 2a

The acceleration of an object in free - fall near the surface of the Earth (neglecting air - resistance) is $a = g=9.8\ m/s^2$ because the only force acting on it is the gravitational force.

Step4: Solve problem 2b

For the apple, $v_0 = 0$, $a = g = 9.8\ m/s^2$, and $t = 1.5\ s$. Using $v=v_0+at$, we have $v = 0+9.8\times1.5$.
$v = 14.7\ m/s$

Step5: Solve problem 2c

The formula for the distance fallen by an object in free - fall is $d=v_0t+\frac{1}{2}at^2$. Since $v_0 = 0$, $a = g = 9.8\ m/s^2$, and $t = 1.5\ s$, we have $d=\frac{1}{2}\times9.8\times(1.5)^2$.
$d=\frac{1}{2}\times9.8\times2.25 = 11.025\ m$

Step6: Solve problem 3a

The acceleration of the rock in free - fall (neglecting air - resistance) is $a = g = 9.8\ m/s^2$.

Step7: Solve problem 3b

For the rock, $v_0 = 0$, $a = g = 9.8\ m/s^2$, and $t = 7\ s$. Using $v=v_0+at$, we get $v=0 + 9.8\times7$.
$v = 68.6\ m/s$

Step8: Solve problem 3c

Using $d=v_0t+\frac{1}{2}at^2$ with $v_0 = 0$, $a = g = 9.8\ m/s^2$, and $t = 7\ s$, we have $d=\frac{1}{2}\times9.8\times7^2$.
$d=\frac{1}{2}\times9.8\times49 = 240.1\ m$

Step9: Solve problem 4a

The acceleration of the rock is $a = g=9.8\ m/s^2$ (downward) as it is in free - fall.

Step10: Solve problem 4b

Given $v_0 = 15\ m/s$, $a = g = 9.8\ m/s^2$, and $t = 5\ s$. Using $v=v_0+at$, we have $v=15 + 9.8\times5$.
$v=15+49=64\ m/s$

Step11: Solve problem 4c

Using $d=v_0t+\frac{1}{2}at^2$ with $v_0 = 15\ m/s$, $a = 9.8\ m/s^2$, and $t = 5\ s$.
$d=15\times5+\frac{1}{2}\times9.8\times5^2=75 + 122.5=197.5\ m$

Step12: Solve problem 5a

The acceleration of the baseball is $a=-g=- 9.8\ m/s^2$ (negative because it acts in the opposite direction of the initial upward motion).

Step13: Solve problem 5b

At the highest point of its motion, the velocity of the baseball is $v = 0\ m/s$.

Step14: Solve problem 5c

When the ball returns to the point from which it was thrown, its velocity has the same magnitude as the initial velocity but is in the opposite direction. So $v=-35\ m/s$.

Step15: Solve problem 6c

At the highest point of the rocket's motion, $v = 0$. Using $v=v_0+at$, we can solve for $t$. Rearranging the formula for $t$ gives $t=\frac{v - v_0}{a}$. Here, $v = 0$, $v_0 = 45\ m/s$, and $a=-g=-9.8\ m/s^2$.
$t=\frac{0 - 45}{-9.8}\approx4.59\ s$

Step16: Solve problem 6d

Using $v^2 - v_0^2=2ad$. At the highest point, $v = 0$, $v_0 = 45\ m/s$, and $a=-9.8\ m/s^2$. Rearranging for $d$ gives $d=\frac{v^2 - v_0^2}{2a}$.
$d=\frac{0 - 45^2}{2\times(-9.8)}=\frac{- 2025}{-19.6}\approx103.32\ m$

Step17: Solve problem 7a

Using $d=v_0t+\frac{1}{2}at^2$, for the acorn, $v_0 = 0$, $a = g = 9.8\ m/s^2$, and $d = 8\ m$. The equation becomes $8=\frac{1}{2}\times9.8\times t^2$. Rearranging for $t$: $t^2=\frac{2\times8}{9.8}$, $t=\sqrt{\frac{16}{9.8}}\approx1.28\ s$.

Step18: Solve problem 7b

Using $v=v_0+at$, with $v_0 = 0$, $a = g = 9.8\ m/s^2$, and $t\approx1.28\ s$. $v=0 + 9.8\times1.28\approx12.54\ m/s$

Step19: Solve problem 8a…

Answer:

  1. $117.6\ m/s$

2a. $9.8\ m/s^2$ (due to gravitational force)
2b. $14.7\ m/s$
2c. $11.025\ m$
3a. $9.8\ m/s^2$
3b. $68.6\ m/s$
3c. $240.1\ m$
4a. $9.8\ m/s^2$
4b. $64\ m/s$
4c. $197.5\ m$
5a. $-9.8\ m/s^2$
5b. $0\ m/s$
5c. $-35\ m/s$
6c. $\approx4.59\ s$
6d. $\approx103.32\ m$
7a. $\approx1.28\ s$
7b. $\approx12.54\ m/s$
8a. $\approx1.28\ m$
8b. $0\ m/s$
8c. $\approx1.02\ s$