Question

c. if you catch the coin at the same height as you released it, how much time did it spend in the air.

1. a rock falls off a cliff that is 100 meters high.

• what is the velocity of the rock when it reaches the ground below the cliff?
• how long did it take the rock to reach the ground?

1. a rocket travelling at +95m/s is accelerated uniformly to +150m/s in 10s. what is the displacement?
2. a tennis ball is thrown straight up with an initial speed of 22.5 m/s. it is caught at the same distance above the ground.

a. how high does the ball rise?
b. how long does the ball remain in the air?

1. suppose an astronaut drops a feather from a height of 1.2 m above the surface of the moon. if the free - fall acceleration on the moon is 1.62 m/s2 downward, how long does it take the feather to hit the moon’s surface?

Explanation:

Step1: For problem 5 - velocity calculation

Use the equation $v^{2}=v_{0}^{2}+2a\Delta y$. Initial velocity $v_{0} = 0$ m/s (starts from rest), acceleration $a = g= 9.8$ m/s², and $\Delta y=100$ m.
$v=\sqrt{v_{0}^{2}+2a\Delta y}=\sqrt{0 + 2\times9.8\times100}=\sqrt{1960}\approx44.3$ m/s

Step2: For problem 5 - time - calculation

Use the equation $\Delta y=v_{0}t+\frac{1}{2}at^{2}$. Since $v_{0} = 0$ m/s, the equation simplifies to $\Delta y=\frac{1}{2}at^{2}$.
$t=\sqrt{\frac{2\Delta y}{a}}=\sqrt{\frac{2\times100}{9.8}}\approx4.52$ s

Step3: For problem 6 - displacement calculation

First, find the acceleration using $a=\frac{v - v_{0}}{t}$, where $v_{0}=95$ m/s, $v = 150$ m/s, and $t = 10$ s. So $a=\frac{150 - 95}{10}=5.5$ m/s². Then use the equation $\Delta x=v_{0}t+\frac{1}{2}at^{2}$.
$\Delta x=95\times10+\frac{1}{2}\times5.5\times10^{2}=950 + 275=1225$ m

Step4: For problem 7a - height calculation

At the maximum - height, the final velocity $v = 0$ m/s. Use the equation $v^{2}=v_{0}^{2}+2a\Delta y$. Here $v_{0}=22.5$ m/s and $a=- 9.8$ m/s².
$\Delta y=\frac{v^{2}-v_{0}^{2}}{2a}=\frac{0 - 22.5^{2}}{2\times(-9.8)}=\frac{- 506.25}{-19.6}\approx25.8$ m

Step5: For problem 7b - time - in - air calculation

Use the equation $v = v_{0}+at$. When the ball returns to the same height, the displacement $\Delta y = 0$. Using $v = v_{0}+at$ and $\Delta y=v_{0}t+\frac{1}{2}at^{2}$, we can also use $v=-v_{0}$ (opposite direction of initial velocity when it comes back to the same height). So $-22.5=22.5-9.8t$.
$9.8t=45$, $t=\frac{45}{9.8}\approx4.59$ s

Step6: For problem 8 - time calculation

Use the equation $\Delta y=v_{0}t+\frac{1}{2}at^{2}$. Since $v_{0} = 0$ m/s (drops from rest), $\Delta y = 1.2$ m and $a = 1.62$ m/s².
$t=\sqrt{\frac{2\Delta y}{a}}=\sqrt{\frac{2\times1.2}{1.62}}=\sqrt{\frac{2.4}{1.62}}\approx1.22$ s

Answer:

• Problem 5 velocity: Approximately $44.3$ m/s
• Problem 5 time: Approximately $4.52$ s
• Problem 6 displacement: $1225$ m
• Problem 7a height: Approximately $25.8$ m
• Problem 7b time in air: Approximately $4.59$ s
• Problem 8 time: Approximately $1.22$ s