Question

1. an arrow leaves a bow at 80m/s. if the arrow is fired straight in the air, how high will the arrow go? (326m)
2. how long will it take the arrow to come back to the ground in the previous question (up and down)? (16sec)
3. at the local swimming hole, there is a 10m cliff (for reference that is about 30ft). someone jumps off the cliff into the lake below. how long will the person be in the air for before they land in the water? (1.4 sec)
4. on the moon an astronaut dropped a rock from 2m high. the rock landed 1.6s later. what is the acceleration due to gravity on the moon? (1.6 m/s²)

Explanation:

Step1: Identify the kinematic - equation for vertical motion

The kinematic equation $v_f^2=v_i^2 - 2gh$ is used for the first question. At the maximum - height, the final velocity $v_f = 0$. The initial velocity $v_i=80m/s$ and the acceleration due to gravity $g = 9.8m/s^2$.
\[0=(80)^2-2\times9.8\times h\]

Step2: Solve for the height $h$

\[2\times9.8\times h=(80)^2\]
\[h=\frac{80^{2}}{2\times9.8}=\frac{6400}{19.6}\approx326.53m\approx326m\]

Step3: For the time of flight of the arrow (question 13)

Use the kinematic equation $v_f=v_i - gt$. When the arrow reaches the maximum - height, $v_f = 0$. First, find the time to reach the maximum - height $t_1$:
\[0 = 80-9.8t_1\]
\[t_1=\frac{80}{9.8}\approx8.16s\]
The total time of flight $t$ (up and down) is $t = 2t_1$. So $t=2\times\frac{80}{9.8}\approx16.33s\approx16s$

Step4: For the person jumping off the cliff (question 14)

Use the kinematic equation $h=v_i t+\frac{1}{2}gt^{2}$. Since the person jumps off the cliff (starts with an initial vertical velocity $v_i = 0$), the equation simplifies to $h=\frac{1}{2}gt^{2}$. Given $h = 10m$ and $g = 9.8m/s^{2}$, we have:
\[10=\frac{1}{2}\times9.8\times t^{2}\]
\[t^{2}=\frac{10\times2}{9.8}=\frac{20}{9.8}\]
\[t=\sqrt{\frac{20}{9.8}}\approx1.43s\approx1.4s\]

Step5: For the acceleration due to gravity on the moon (question 15)

Use the kinematic equation $h=v_i t+\frac{1}{2}a t^{2}$. Since the rock is dropped ($v_i = 0$), $h = 2m$ and $t = 1.6s$.
\[2=\frac{1}{2}\times a\times(1.6)^{2}\]
\[a=\frac{2\times2}{(1.6)^{2}}=\frac{4}{2.56}=1.5625m/s^{2}\approx1.6m/s^{2}\]

Answer:

1. $326m$
2. $16s$
3. $1.4s$
4. $1.6m/s^{2}$