Question

exercises: 1) a diver jumps from a cliff and is in the air for 4 seconds before she reaches the water below. what is the displacement of the diver? δx = v₀(t) + ½(a)(t) δx = 0(4) + ½(-9.8)(4) -19.6m = δx 2) a ball is thrown from a bridge directly upward with a speed of 15 m/s. the ball goes up, falls down, and then hits the ground 10 seconds later. calculate the displacement of the ball. δx = v₀(t) + ½(a)(t) δx = 15(10) + ½(-9.8)(10) 150 - 49 = 101m = δx 3) a ball is dropped from a bridge that has a height of 45m tall. if the ball is released from rest, calculate the time it will take the ball to reach the bottom of the bridge. 45 m = 0(t) + ½(-9.8)(t)

Explanation:

Step1: Identify the kinematic - equation

The kinematic equation for displacement is $\Delta x = v_0t+\frac{1}{2}at^{2}$, where $\Delta x$ is displacement, $v_0$ is initial velocity, $t$ is time, and $a$ is acceleration. The acceleration due to gravity $a=- 9.8m/s^{2}$.

Step2: Solve for the diver's displacement

The diver jumps from rest, so $v_0 = 0m/s$. Using the kinematic - equation $\Delta x=v_0t+\frac{1}{2}at^{2}$, substituting $v_0 = 0m/s$, $t = 4s$, and $a=-9.8m/s^{2}$:
$\Delta x=0\times4+\frac{1}{2}\times(-9.8)\times4^{2}$
$\Delta x=-78.4m$. The negative sign indicates the direction is downward.

Step3: Solve for the ball's displacement

The ball is thrown upward with $v_0 = 15m/s$ and $t = 10s$. Using the kinematic - equation $\Delta x=v_0t+\frac{1}{2}at^{2}$, substituting $v_0 = 15m/s$, $t = 10s$, and $a=-9.8m/s^{2}$:
$\Delta x=15\times10+\frac{1}{2}\times(-9.8)\times10^{2}$
$\Delta x = 150-490=-340m$. There is an error in the original work for this part. The correct calculation is $\Delta x=15\times10+\frac{1}{2}\times(-9.8)\times10^{2}=150 - 490=-340m$. The negative sign indicates the ball ends up below its starting point.

Step4: Solve for the time of the dropped ball

The ball is dropped from rest ($v_0 = 0m/s$), and the height of the bridge $h=\Delta x = 45m$ (taking downwards as positive for this part). Using the kinematic - equation $\Delta x=v_0t+\frac{1}{2}at^{2}$, substituting $v_0 = 0m/s$ and $\Delta x = 45m$, $a = 9.8m/s^{2}$ (taking downwards as positive), we get $45=0\times t+\frac{1}{2}\times9.8\times t^{2}$.
$4.9t^{2}=45$, then $t^{2}=\frac{45}{4.9}\approx9.18$, and $t=\sqrt{9.18}\approx3.03s$.

Answer:

1. The diver's displacement is $-78.4m$.
2. The ball's displacement is $-340m$.
3. The time for the ball to reach the bottom is approximately $3.03s$.