Question

Nadia threw a 4.1 kg ball straight upward with a speed of 15m/s. How high is the ball at the very top just before it starts falling back down

Answer 1

11.48 meters

Explanation:

Let's establish the equation of the situation:

`(1)/(2)\cdot m\cdot(v_i)^2+m\cdot g\cdot h_i=(1)/(2)\cdot m\cdot(v_f)^2+m\cdot g\cdot h_f`

We can determine that according to the situation the initial speed is 15 m/s, the final height is 0, the final speed is 0, therefore, we must calculate the final height, like this:

Replacing tue values:

`\begin{gathered} (1)/(2)\cdot1\cdot15^2+1\cdot9.8\cdot\cdot0=(1)/(2)\cdot1\cdot0+1\cdot9.8\cdot\: h_f \\ 112.5+0=0+9.8\cdot h_f \\ h_f=(112.5)/(9.8) \\ h_f=11.48\text{ m} \end{gathered}`

Therefore, the height is 11.48 meters