Question

A 74.1 kg high jumper leaves the ground with

Answer 1

Answer:

4.80 m

Explanation:

We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.

Let's set the upwards direction to be positive and the downwards direction to be negative.

List out the relevant known variables.

1. v₀ = 9.7 m/s
2. a = -9.8 m/s²
3. Δx = ?

We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.

    4. v = 0 m/s

Using these four variables, let's find the constant acceleration equation that contains these variables:

• v² = v₀² + 2aΔx

Substitute the known values into the equation and solve for Δx.

• (0)² = (9.7)² + 2(-9.8)Δx
• 0 = 94.09 + (-19.6)Δx
• -94.09 = -19.6Δx
• Δx = 4.80

The high jumper can jump to a max height of 4.80 m.