Question

1. a rocket is launched from the ground with an initial velocity of 25 miles per hour. when the rocket reaches its highest point, it has a velocity of zero for a split second. what is the height of the rocket in meters when it reaches its highest point? 0.6 m 12.77 m 19 m 6.4 m

Explanation:

Step1: Convert initial velocity units

We know that \(1\) mile \( = 1609.34\) meters and \(1\) hour \(=3600\) seconds.
The initial velocity \(v_0 = 25\) miles per hour.

$$v_0=25\times\frac{1609.34}{3600}\text{ m/s}\approx11.176\text{ m/s}$$

Step2: Use the kinematic equation \(v^{2}-v_{0}^{2}=- 2gh\)

At the highest - point, \(v = 0\) (velocity is zero). The kinematic equation \(v^{2}-v_{0}^{2}=-2gh\) (where \(g = 9.8\text{ m/s}^2\) is the acceleration due to gravity and \(h\) is the height).
Substitute \(v = 0\) and \(v_0\approx11.176\text{ m/s}\) into the equation:

$$0-(11.176)^{2}=-2\times9.8\times h$$

$$h=\frac{(11.176)^{2}}{2\times9.8}$$

$$h=\frac{124.9}{19.6}\approx6.4\text{ m}$$

Answer:

6.4 m