Question

1. a stone is dropped from rest from the top of a cliff. you hear it hit the bottom 3 seconds after you drop the stone. how high is the cliff? what will the velocity of the stone be at the bottom of the well? (44.1m, -29.4m/s)
2. a ball is thrown downward at a velocity of 10m/s from the top of a building. it took the ball 2 seconds to land. how tall is the building? what will the velocity of the ball be at the bottom? (40m, -29.6m/s)
3. a car traveling 20m/s east slams on the brakes and skids to a halt. the mark on the road left by the skid is 50m long. what was the cars acceleration? how long did it take for the car to stop? (-4m/s², 5seconds)
4. a car was traveling at 30m/s east and accelerated to 50m/s east. the cars displacement during this acceleration was 400m. what was the cars acceleration? (2m/s² east)

Explanation:

Problem 21

Step1: Find height of cliff

The stone is in free - fall, and we use the equation $h = v_0t+\frac{1}{2}gt^2$. Since it is dropped from rest, $v_0 = 0\ m/s$, and $g=- 9.8\ m/s^2$, $t = 3\ s$.
$h=0\times t+\frac{1}{2}\times(-9.8)\times3^2=\frac{-9.8\times9}{2}=- 44.1\ m$. The height of the cliff is $44.1\ m$ (the negative sign just indicates the direction is downwards).

Step2: Find velocity at the bottom

We use the equation $v = v_0+gt$. With $v_0 = 0\ m/s$, $g=-9.8\ m/s^2$ and $t = 3\ s$.
$v=0+( - 9.8)\times3=-29.4\ m/s$

Step1: Find height of building

We use the equation $h = v_0t+\frac{1}{2}gt^2$, where $v_0=-10\ m/s$ (downward direction), $g = - 9.8\ m/s^2$ and $t = 2\ s$.
$h=(-10)\times2+\frac{1}{2}\times(-9.8)\times2^2=-20 - 19.6=-40\ m$. The height of the building is $40\ m$.

Step2: Find velocity at the bottom

We use the equation $v = v_0+gt$. Substituting $v_0=-10\ m/s$, $g=-9.8\ m/s^2$ and $t = 2\ s$.
$v=-10+( - 9.8)\times2=-10-19.6=-29.6\ m/s$

Step1: Find acceleration

We use the equation $v^2=v_0^2 + 2ax$. The car stops, so $v = 0\ m/s$, $v_0 = 20\ m/s$ and $x = 50\ m$.
$0=(20)^2+2a\times50$. Then $0 = 400+100a$, and $100a=-400$, so $a=-4\ m/s^2$.

Step2: Find time to stop

We use the equation $v = v_0+at$. With $v = 0\ m/s$, $v_0 = 20\ m/s$ and $a=-4\ m/s^2$.
$0 = 20+( - 4)t$, so $4t = 20$ and $t = 5\ s$

Answer:

Height of cliff: $44.1\ m$, Velocity at bottom: $-29.4\ m/s$

Problem 22