Question

question 18 (1 point)
a ball is thrown vertically upward with a speed v. an identical second ball is thrown upward with a speed 2v (twice as fast). what is the ratio of the maximum height of the second ball to that of the first ball? (how many times higher does the second ball go than the first ball?)
4:1
2:1
1.7:1
1.4:1

Explanation:

Step1: Use the kinematic equation

The kinematic equation \(v^{2}=v_{0}^{2}-2gh\) (at maximum height \(v = 0\)). For the first ball, \(0 = v^{2}-2gh_{1}\), so \(h_{1}=\frac{v^{2}}{2g}\). For the second ball, \(0=(2v)^{2}-2gh_{2}\), so \(h_{2}=\frac{(2v)^{2}}{2g}=\frac{4v^{2}}{2g}\).

Step2: Calculate the ratio

The ratio \(\frac{h_{2}}{h_{1}}=\frac{\frac{4v^{2}}{2g}}{\frac{v^{2}}{2g}}\). The \(\frac{v^{2}}{2g}\) terms cancel out, and \(\frac{h_{2}}{h_{1}} = 4\).

Answer:

A. \(4:1\)