Question

a 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. if the initial magnitude of the velocity of the ball is 20.0 m/s. what is the magnitude of the final velocity when it strikes the ground? (note: ignore air resistance.)
a 12.7 m/s
b 34.9 m/s
c 5.2 m/s
d 28.1 m/s

Explanation:

Step1: Apply the conservation of mechanical energy

The initial mechanical energy \(E_{i}\) is the sum of kinetic energy \(K_{i}=\frac{1}{2}mv_{i}^{2}\) and potential energy \(U_{i} = mgh\). The final mechanical energy \(E_{f}\) is just kinetic energy \(K_{f}=\frac{1}{2}mv_{f}^{2}\) (since \(h = 0\) at the ground). According to the conservation of mechanical energy \(E_{i}=E_{f}\), so \(\frac{1}{2}mv_{i}^{2}+mgh=\frac{1}{2}mv_{f}^{2}\).

Step2: Solve for \(v_{f}\)

Cancel out the mass \(m\) from each term of the equation \(\frac{1}{2}v_{i}^{2}+gh=\frac{1}{2}v_{f}^{2}\). Then \(v_{f}=\sqrt{v_{i}^{2} + 2gh}\). Substitute \(v_{i}=20.0\space m/s\), \(g = 9.8\space m/s^{2}\), and \(h=20.0\space m\) into the formula: \(v_{f}=\sqrt{(20.0)^{2}+2\times9.8\times20.0}=\sqrt{400 + 392}=\sqrt{792}\approx28.1\space m/s\)

Answer:

d. \(28.1\space m/s\)