Janet jumps off a high diving platform with
a horizontal velocity of 2.37 m/s and lands in
the water 1.9 s later.
How high is the platform? The acceleration
of gravity is 9.8 m/s
2
.
Answer in units of m
The horizontal velocity is unimportant. No one is asking how far away from the platform she lands.
So, all we want to know is how far one falls in 1.9s under gravity.
s = .5at^2 = 4.9*1.9^2 = 17.7m
That's a nice high platform!
To find the height of the diving platform, we can use the equations of motion under constant acceleration.
First, let's list out the information given:
- Initial horizontal velocity, v₀x = 2.37 m/s
- Time of flight, t = 1.9 s
- Acceleration due to gravity, g = 9.8 m/s²
Since the platform and the landing point are at the same horizontal level, the vertical motion is the only one that affects the height. We can use the equation:
Δy = v₀y * t + 0.5 * g * t²
To apply this equation, we need to determine the initial vertical velocity, v₀y. Given that there is no initial vertical velocity (she jumps vertically downwards), we have v₀y = 0 m/s.
Using this information, the equation becomes:
Δy = 0 * t + 0.5 * g * t²
Δy = 0 + 0.5 * 9.8 * (1.9)²
Simplifying the equation:
Δy = 0.5 * 9.8 * 3.61
Δy = 17.669 m
Therefore, the height of the diving platform is approximately 17.669 m.