A lowly high diver pushes off horizontally with a speed of 2.75 m/s from the edge of a platform that is 10.0 m above the surface of the water. (a) At what horizontal distance from the edge of the platform is the diver 0.866 s after pushing off? (b) At what vertical distance above the surface of the water is the diver just then? (c) At what horizontal distance from the edge of the platform does the diver strike the water?
a. d = Xo * t = 2.75m/s * 0.866s=2.38 m.
b, d=0.5g*t^2 = 4.9*(0.866)^2 = 3.67 m.
c. d = 0.5g*t^2 = 10 m.
4.9t^2 = 10
t^2 = 2.04
Tf = 1.428 s. = Fall time.
D = Xo * Tf = 2.75 * 1.428 = 3.93 m.
To solve this problem, we can use the equations of motion for horizontal and vertical motion separately. Let's break it down step by step:
(a) To find the horizontal distance from the edge of the platform after 0.866 s, we need to use the horizontal velocity. The horizontal motion is unaffected by gravity, so the diver will continue moving with a constant horizontal velocity.
The equation we can use is:
Horizontal distance = Horizontal velocity × Time
Given:
Initial horizontal velocity (v) = 2.75 m/s
Time (t) = 0.866 s
Plugging these values into the equation, we get:
Horizontal distance = 2.75 m/s × 0.866 s
Horizontal distance ≈ 2.38 meters
Therefore, the diver is approximately 2.38 meters away from the edge of the platform after 0.866 s.
(b) To find the vertical distance above the water surface at that time, we can use the equation of motion for vertical motion.
The equation we can use is:
Vertical distance = Initial vertical velocity × Time + (1/2) × Acceleration due to gravity × Time²
Given:
Initial vertical velocity (u) = 0 m/s (since the diver is only pushing off horizontally)
Time (t) = 0.866 s
Acceleration due to gravity (g) = 9.8 m/s²
Plugging these values into the equation, we get:
Vertical distance = 0 m/s × 0.866 s + (1/2) × 9.8 m/s² × (0.866 s)²
Vertical distance ≈ 3.84 meters
Therefore, the diver is approximately 3.84 meters above the surface of the water after 0.866 s.
(c) To find the horizontal distance from the edge of the platform where the diver strikes the water, we need to consider the time it takes for the diver to fall from the platform to the water.
The equation we can use is:
Vertical distance = Initial vertical velocity × Time + (1/2) × Acceleration due to gravity × Time²
Given:
Initial vertical velocity (u) = 0 m/s (since the diver is only pushing off horizontally)
Vertical distance = 10.0 m (height of platform)
Acceleration due to gravity (g) = 9.8 m/s²
We need to solve this equation to find the time it takes for the diver to fall:
10.0 m = 0 m/s × Time + (1/2) × 9.8 m/s² × Time²
Simplifying the equation, we get:
4.9 Time² = 10.0
Time² ≈ 2.04
Time ≈ √2.04 ≈ 1.43 s
Now we can use the horizontal velocity and the time calculated to find the horizontal distance:
Horizontal distance = Horizontal velocity × Time
Given:
Horizontal velocity = 2.75 m/s
Time = 1.43 s
Plugging these values into the equation, we get:
Horizontal distance = 2.75 m/s × 1.43 s
Horizontal distance ≈ 3.93 meters
Therefore, the diver strikes the water at a horizontal distance of approximately 3.93 meters from the edge of the platform.
To answer these questions, we can use the equations of motion for projectile motion. Let's break down each question and determine the steps involved in finding the answers:
(a) At what horizontal distance from the edge of the platform is the diver 0.866 s after pushing off?
To find the horizontal distance, we need to use the formula for horizontal motion:
Horizontal distance (d) = initial horizontal velocity (v₀x) × time (t)
In this case, the initial horizontal velocity is given as 2.75 m/s, and the time is 0.866 s. Substitute these values into the formula to find the horizontal distance.
(b) At what vertical distance above the surface of the water is the diver just then?
The vertical distance can be determined using the formula for vertical motion:
Vertical distance (h) = initial vertical velocity (v₀y) × time (t) + 0.5 × acceleration due to gravity (g) × time²
The initial vertical velocity can be determined using the formula: v₀y = initial speed × sin(θ), where θ is the angle of projection. In this case, since the diver pushes off horizontally, the angle of projection is 0 degrees, so sin(0) = 0.
Therefore, the initial vertical velocity (v₀y) is 0 m/s.
The acceleration due to gravity (g) is approximately 9.8 m/s².
(c) At what horizontal distance from the edge of the platform does the diver strike the water?
To find the horizontal distance when the diver strikes the water, we need to use the same formula as in part (a), but this time we will plug in the total time of flight. The time of flight can be determined by doubling the time taken to reach the maximum height since the dive is symmetrical.
These steps should help you solve the problem.