a diver makes a running horizontal jump off a cliff located 4m above the water level below. how fast must the driver be traveling at launch in order to clear 6m section of rocks at the water's edge?
well, how long does it take to fall 4m?
divide that into 6m to get the speed.
To find the speed the diver must be traveling at launch, we can use the concept of projectile motion. In this case, the horizontal motion is independent of the vertical motion.
Given:
- Height of the cliff (h): 4m
- Horizontal distance to clear (d): 6m
We need to determine the launch speed (v).
First, let's find the time it takes for the diver to fall from the top of the cliff to the water level. We can use the equation for vertical motion:
h = (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time of flight.
Replacing the values, the equation becomes:
4 = (1/2) * 9.8 * t^2
Simplifying,
8 = 4.9 * t^2
Dividing both sides by 4.9,
t^2 = 1.63
Taking the square root on both sides,
t ≈ 1.28 seconds
We can now calculate the horizontal velocity (v) using the formula:
d = v * t
Rearranging the equation to solve for v:
v = d / t
Substituting the given values,
v = 6 / 1.28
v ≈ 4.69 m/s
Therefore, the diver must be traveling at a speed of approximately 4.69 m/s at launch to clear the 6m section of rocks at the water's edge.
To calculate the required launch speed for the diver to clear the 6m section of rocks at the water's edge, we can use the principles of projectile motion.
We know that the horizontal motion of the diver is not affected by gravity because there are no horizontal forces acting on the diver. Therefore, we only need to consider the vertical motion.
Let's break down the problem into two parts:
1. The vertical distance from the cliff to the water's edge:
The diver clears a vertical distance equal to the sum of the height of the cliff (4m) and the height of the rocks (6m): 4m + 6m = 10m.
2. The horizontal distance the diver travels:
Since there are no horizontal forces acting on the diver, the horizontal distance is simply the horizontal speed multiplied by the time of flight. We can use the horizontal distance formula: distance = speed × time.
Now, to find the required launch speed, we need to calculate the time of flight.
1. Vertical motion:
We can use the equation of motion for vertical motion:
Δy = V_initial * t - 0.5 * g * t^2
where:
Δy = vertical distance (10m)
V_initial = initial vertical velocity (which is 0 for our case because the diver jumps horizontally)
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight
Rearranging the equation, we get:
0.5 * g * t^2 = V_initial * t - Δy
0.5 * 9.8 * t^2 = 0 - 10
4.9 * t^2 = 10
t^2 = 10 / 4.9
t ≈ √2.04
t ≈ 1.43 seconds (rounded to two decimal places)
2. Horizontal motion:
We want to find the launch speed, so we need to calculate the horizontal distance. We know that the horizontal distance is equal to the horizontal speed multiplied by the time of flight.
Using the distance formula: distance = speed × time
The horizontal distance = 6m (given in the question)
6 = speed * 1.43
speed ≈ 4.20 m/s (rounded to two decimal places)
Therefore, the diver must be traveling at approximately 4.20 m/s at launch in order to clear the 6m section of rocks at the water's edge.