Question

1. you throw a ball straight into the air. the ball landed 2 seconds later. how high did the ball go? what velocity did you throw the ball? what was the velocity of the ball when it hit the ground? (4.9m, 9.8m/s, -9.8m/s)
2. calculate the \hang time\ of a vertical jumper who jumps a vertical distance of .75m. (.78s)
3. you plan to measure your reaction time by dropping a ruler and catching it. you start your fingers at the 0cm line and catch the ruler at 15cm. what is your reaction time? (.17s)
4. you throw a ball upward at 25m/s. what is the velocity of the ball when it is 30m high? (about 6m/s upward)

Explanation:

Question 25

Step1: Find initial velocity

The time of flight $t = 2s$. The time to reach the maximum - height is $t_{up}=\frac{t}{2}=1s$. Using the equation $v = v_0−gt$ (at maximum - height $v = 0$), and $g = 9.8m/s^2$. So, $0=v_0 - 9.8\times1$, then $v_0=9.8m/s$.

Step2: Calculate maximum height

Use the equation $v^{2}-v_{0}^{2}=-2gh$. At maximum - height $v = 0$ and $v_0 = 9.8m/s$. So, $0-(9.8)^{2}=-2\times9.8\times h$, solving for $h$ gives $h = 4.9m$.

Step3: Determine final velocity

Using the equation $v = v_0−gt$. Here, $v_0 = 9.8m/s$, $t = 2s$, and $g = 9.8m/s^2$. So, $v=9.8-9.8\times2=-9.8m/s$.

Step1: Use the kinematic equation

The kinematic equation for vertical motion is $h = v_0t-\frac{1}{2}gt^{2}$. At the maximum - height, the final velocity $v = v_0 - gt = 0$, so $v_0=gt$. Substituting $v_0$ into $h = v_0t-\frac{1}{2}gt^{2}$ gives $h=gt\times t-\frac{1}{2}gt^{2}=\frac{1}{2}gt^{2}$.

Step2: Solve for time

We know $h = 0.75m$ and $g = 9.8m/s^2$. From $h=\frac{1}{2}gt^{2}$, we have $t=\sqrt{\frac{2h}{g}}$. Plugging in the values: $t=\sqrt{\frac{2\times0.75}{9.8}}\approx0.39s$. The hang - time $T = 2t\approx0.78s$.

Step1: Use the free - fall equation

The ruler is in free - fall, and the equation for free - fall displacement is $h=\frac{1}{2}gt^{2}$, where $h = 15cm=0.15m$ and $g = 9.8m/s^2$.

Step2: Solve for reaction time

From $h=\frac{1}{2}gt^{2}$, we can solve for $t$: $t=\sqrt{\frac{2h}{g}}$. Plugging in $h = 0.15m$ and $g = 9.8m/s^2$ gives $t=\sqrt{\frac{2\times0.15}{9.8}}\approx0.17s$.

Answer:

The ball goes $4.9m$ high. The initial velocity is $9.8m/s$. The final velocity when it hits the ground is $-9.8m/s$.

Question 26