Question

A diver leaves a 3-m

board on a trajectory that takes her 2.8 m
above the board and then into the water 2.7 m
horizontally from the end of the board.
At what speed did she leave the board?
Express your answer to two significant figures and include the appropriate units.

Answer 1

4.1 m/s (two sig figs)

Explanation:

Honestly, this might have been more physics but oh well.

Given:

Vertical displacement (y) = 2.8 meters

Horizontal displacement (x) = 2.7 meters

Acceleration due to gravity (g) is approximately 9.81 m/s²

Vertical Component:

Using the equation for vertical displacement:

y = v_0y * t - 0.5 * g * t^2

When reaching the highest point (half of the total time), y = 1.4 meters:

1.4 = v_0y * (t / 2) - 0.5 * g * (t / 2)^2

Horizontal Component:

Using the equation for horizontal displacement:

x = v_0x * t

Connecting the Components:

From the horizontal equation, solve for t:

t = x / v_0x

Substitute this t value into the vertical equation:

1.4 = v_0y * (x / v_0x) - 0.5 * g * (x / v_0x)^2

Solve this equation for v_0x, the speed at which she leaves the board.

Rounded to two significant figures, the speed is approximately 4.1 m/s.