Question

3.1. A stone of mass 2kg is thrown vertically upward with a speed

of 20m's Calculate the maximum height reached.​

Answer 1

20.41 m

Step-by-step explanation:

Let's list out the relevant variables in this problem. Take the upwards direction to be positive and the downwards to be negative.

• v₀ = 20 m/s
• v = 0 m/s
• a = -9.8 m/s²
• Δy = ?

We are using the fact that at maximum height, the object's velocity is 0 m/s right before it changes direction to fall vertically back down to Earth.

Find the constant acceleration equation that contains these four variables.

• v² = v₀² + 2aΔy

Substitute the known values into the equation and solve for Δy, the maximum height.

• (0)² = (20)² + 2(-9.8)Δy
• 0 = 400 - 19.6Δy
• -400 = -19.6Δy
• Δy = 20.4081632653

The maximum height of the stone is 20.41 m. Notice that its mass was irrelevant when solving for its maximum height when thrown.