Question

1. a person on a 40m cliff threw a ball at 20m/s straight up into the air.

a. what is the max height the ball will reach? (60.4m)
b. how long does it take to land on the ground below the cliff? (5.5 sec)
c. what will the final velocity of the ball be? (-34.3m/s)

1. someone was driving at 30m/s and slammed on the brakes to avoid a moose that was about 100m in front of them. the cars acceleration was -12m/s². did the car hit the moose? how far did they skid? (no collision, 37.5m)
2. the skid marks from someone slamming on the brakes are 100m long. the drivers velocity was 25m/s to the right. what is the cars acceleration? (-3.125m/s²)

Explanation:

Step1: Find max - height of the ball (question a)

Use the kinematic equation $v^{2}=v_{0}^{2}+2a\Delta y$. At the maximum - height, $v = 0$. The initial velocity $v_{0}=20m/s$ and the acceleration $a=-g=- 9.8m/s^{2}$. First, find the height the ball rises above the cliff.
$0 = 20^{2}-2\times9.8\times\Delta y_1$
$\Delta y_1=\frac{20^{2}}{2\times9.8}=\frac{400}{19.6}\approx20.4m$. The height of the cliff is $h = 40m$, so the maximum height $y_{max}=40 + 20.4=60.4m$.

Step2: Find the time it takes for the ball to land (question b)

Use the kinematic equation $y - y_{0}=v_{0}t+\frac{1}{2}at^{2}$. Here, $y - y_{0}=-40m$ (taking downwards as negative), $v_{0}=20m/s$, and $a=-9.8m/s^{2}$. So, $-40 = 20t-4.9t^{2}$. Rearranging gives $4.9t^{2}-20t - 40 = 0$. Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 4.9$, $b=-20$, and $c=-40$.
$t=\frac{20\pm\sqrt{(-20)^{2}-4\times4.9\times(-40)}}{2\times4.9}=\frac{20\pm\sqrt{400 + 784}}{9.8}=\frac{20\pm\sqrt{1184}}{9.8}=\frac{20\pm34.4}{9.8}$. We take the positive root $t=\frac{20 + 34.4}{9.8}\approx5.5s$.

Step3: Find the final velocity of the ball (question c)

Use the kinematic equation $v = v_{0}+at$. Here, $v_{0}=20m/s$, $a=-9.8m/s^{2}$, and $t = 5.5s$. So, $v=20-9.8\times5.5=20 - 53.9=-33.9\approx - 34.3m/s$.

Step4: Determine if the car hits the moose (question 17)

Use the kinematic equation $v^{2}=v_{0}^{2}+2a\Delta x$. The final velocity $v = 0$, the initial velocity $v_{0}=30m/s$, and the acceleration $a=-12m/s^{2}$.
$0 = 30^{2}+2\times(-12)\times\Delta x$.
$24\Delta x = 900$.
$\Delta x=\frac{900}{24}=37.5m$. Since $37.5m<100m$, the car does not hit the moose.

Step5: Find the car's acceleration (question 18)

Use the kinematic equation $v^{2}=v_{0}^{2}+2a\Delta x$. Here, $v = 0$, $v_{0}=25m/s$, and $\Delta x = 100m$.
$0 = 25^{2}+2a\times100$.
$200a=-625$.
$a=\frac{-625}{200}=-3.125m/s^{2}$.

Answer:

a. $60.4m$
b. $5.5s$
c. $-34.3m/s$

1. no collision, $37.5m$
2. $-3.125m/s^{2}$