A runner runs and jumps off a cliff into the water below. If the runner takes up running a 9.37 m/s and lands in the water 2.0 m away from the cliff's edge: How tall is the cliff?
The time it takes to hit the water is
t = 2.0/9.37 = 0.213 s
Now calculate how far he falls in that time.
Y = (g/2) t^2
That will be the cliff height.
To determine the height of the cliff, we can use the principles of projectile motion. The horizontal velocity of the runner, 9.37 m/s, remains constant throughout the jump. The vertical motion can be divided into two parts: upward motion and downward motion.
First, let's determine the time it takes for the runner to travel horizontally 2.0 m. We can use the formula:
distance = velocity * time
In this case, the distance is 2.0 m and the velocity is 9.37 m/s. Rearranging the formula gives us:
time = distance / velocity
time = 2.0 m / 9.37 m/s
time ≈ 0.2135 s
Next, we will use this time to calculate the height of the cliff. During the time of flight, the vertical motion can be analyzed using the following formula:
vertical distance = initial vertical velocity * time + 0.5 * acceleration * time^2
The initial vertical velocity is 0 m/s because the runner starts from rest in the vertical direction. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
Since the runner jumps vertically upwards and then falls downwards, the total time of flight is twice the time calculated previously:
total time = 2 * 0.2135 s
total time ≈ 0.427 s
Now, we can calculate the height of the cliff:
vertical distance = 0 * 0.427 s + 0.5 * 9.8 m/s^2 * (0.427 s)^2
vertical distance ≈ 0 + 0.8861 m
vertical distance ≈ 0.8861 m
Therefore, the height of the cliff is approximately 0.8861 meters.