Question

rick shoots a basketball at an angle of 35° from the horizontal. it leaves his hands 7 feet from the ground with a velocity of 20 ft/s.
step 2 of 2: determine the height of the ball when it is 14 feet away horizontally. round to the nearest tenth.

Explanation:

Step1: Find the time when the horizontal distance is 14 feet

The horizontal component of the velocity \(v_{0x}=v_0\cos\theta\), where \(v_0 = 20\) ft/s and \(\theta = 35^{\circ}\). So \(v_{0x}=20\cos35^{\circ}\approx20\times0.8192 = 16.384\) ft/s.
Using the formula \(x = v_{0x}t\), we solve for \(t\) when \(x = 14\) feet. So \(t=\frac{x}{v_{0x}}=\frac{14}{16.384}\approx0.854\) s.

Step2: Find the height of the ball at time \(t\)

The vertical - motion formula is \(y=y_0+v_{0y}t-\frac{1}{2}gt^2\), where \(y_0 = 7\) feet, \(v_{0y}=v_0\sin\theta=20\sin35^{\circ}\approx20\times0.5736 = 11.472\) ft/s, and \(g = 32\) ft/s².
Substitute \(t = 0.854\) s into the formula:

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Answer:

\(5.1\) feet