Question

1. a ball is thrown downwards at -3 m/s. how long does it take for the ball to reach a velocity of -12.8 m/s? show your work below.
2. a student drops a coin off of a cliff. it takes 4.5 seconds for the coin to hit the ground below. what must be the height of the cliff? show all of your work below and include units.
3. a student throws a dart upwards with some unknown initial velocity. it takes the dart a total of 2.7 seconds to go up to its maximum height and come back down to where it started.

a. what must be the displacement of the dart? briefly explain your answer.
b. calculate the initial velocity of the dart. show all of your work below and remember to include units.

Explanation:

Step1: Identify the kinematic - equation for question 4

The equation $v = v_0+at$ is used, where $v$ is final velocity, $v_0$ is initial velocity and $a$ is acceleration. In free - fall, $a=- 9.8\ m/s^2$.
$v = v_0+at$

Step2: Rearrange the equation to solve for $t$

$t=\frac{v - v_0}{a}$. Substitute $v=-12.8\ m/s$, $v_0 = - 3\ m/s$ and $a=-9.8\ m/s^2$ into the formula.
$t=\frac{-12.8-(-3)}{-9.8}=\frac{-12.8 + 3}{-9.8}=\frac{-9.8}{-9.8}=1\ s$

Step3: Identify the kinematic - equation for question 5

The equation $y=v_0t+\frac{1}{2}at^2$ is used. Since the coin is dropped, $v_0 = 0\ m/s$, $a=-9.8\ m/s^2$ and $t = 4.5\ s$.
$y=v_0t+\frac{1}{2}at^2$

Step4: Calculate the height of the cliff

Substitute the values into the formula: $y=0\times4.5+\frac{1}{2}\times(-9.8)\times(4.5)^2=- \frac{1}{2}\times9.8\times20.25=-99.225\ m$. The negative sign indicates the direction is downwards, and the height of the cliff is $99.225\ m$.

Step5: Answer question 6a

Displacement is the change in position from the starting point to the ending point. Since the dart returns to its starting position, the displacement $\Delta x=0\ m$.

Step6: Identify the kinematic - equation for question 6b

The time to reach the maximum height is half of the total time of flight. So $t=\frac{2.7}{2}=1.35\ s$. At the maximum height, the final velocity $v = 0\ m/s$. Using the equation $v = v_0+at$, we can solve for $v_0$.
$v = v_0+at$

Step7: Rearrange the equation to solve for $v_0$

$v_0=v - at$. Substitute $v = 0\ m/s$, $a=-9.8\ m/s^2$ and $t = 1.35\ s$ into the formula.
$v_0=0-(-9.8)\times1.35=13.23\ m/s$

Answer:

1. $t = 1\ s$
2. The height of the cliff is $99.225\ m$

6a. The displacement of the dart is $0\ m$
6b. The initial velocity of the dart is $13.23\ m/s$