Question

dana shoots a basketball at an angle of 45° from the horizontal. it leaves her hands 6 feet from the ground with a velocity of 27 ft/s. step 2 of 2: determine the height of the ball when it is 27 feet away horizontally. round to the nearest tenth.

Explanation:

Step1: Find the time taken to reach 27 feet horizontally

The horizontal component of velocity \(v_{x}=v\cos\theta\), where \(v = 27\) ft/s and \(\theta=45^{\circ}\), so \(v_{x}=27\cos45^{\circ}=27\times\frac{\sqrt{2}}{2}\approx19.09\) ft/s.
Using the formula \(x = v_{x}t\), where \(x = 27\) ft. Then \(t=\frac{x}{v_{x}}=\frac{27}{27\cos45^{\circ}}=\sqrt{2}\approx1.414\) s.

Step2: Find the height of the ball

The vertical - motion equation is \(y=y_{0}+v_{y}t-\frac{1}{2}gt^{2}\), where \(y_{0} = 6\) ft, \(v_{y}=v\sin\theta=27\sin45^{\circ}\approx19.09\) ft/s, and \(g = 32\) ft/s².
Substitute \(t=\sqrt{2}\) into the equation:

$$
LATEXBLOCK0
$$

Answer:

\(1.0\)