a girl jumps off a high diving platform with a horizontal velocity of 2.77m/s and lands in the water 1.2s later the acceleration of gravity is 9.8m/s squared
how high is the platform and how far from the base of the platform does she land?
first do the vertical problem
We will have to assume she has no initial speed up, just horizontal
h = h initial + Vi t - 4.9 t^2
0 = h initial + 0 (1.2) - 4.9 (1.2)^2
solve for h initial
her horizontal speed never changed so horizontal distance is 2.77*1.
To determine the height of the platform and the horizontal distance the girl lands from the base of the platform, we can use the following kinematic equations:
For vertical motion:
1. Vertical displacement (Δy) = initial vertical velocity (Vy) * time (t) + 0.5 * acceleration due to gravity (g) * (time (t))^2
2. Vertical velocity (Vy) = acceleration due to gravity (g) * time (t)
For horizontal motion:
3. Horizontal displacement (Δx) = horizontal velocity (Vx) * time (t)
Given:
- Initial horizontal velocity (Vx) = 2.77 m/s
- Time (t) = 1.2 s
- Acceleration due to gravity (g) = 9.8 m/s^2
1. Calculate vertical displacement (Δy):
Using equation 1,
Δy = 0 * 1.2 + 0.5 * 9.8 * (1.2)^2
Δy = 7.056 m
2. Calculate vertical velocity (Vy):
Using equation 2,
Vy = 9.8 * 1.2
Vy = 11.76 m/s
3. Calculate horizontal displacement (Δx):
Using equation 3,
Δx = 2.77 * 1.2
Δx = 3.324 m
Therefore, the height of the platform is approximately 7.056 meters, and the girl lands approximately 3.324 meters from the base of the platform.