rick shoots a basketball at an angle of 35° from the horizontal. it leaves his hands 7 feet from the ground…

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.\nstep 2 of 2: determine the height of the ball when it is 14 feet away horizontally. round to the nearest tenth.

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.\nstep 2 of 2: determine the height of the ball when it is 14 feet away horizontally. round to the nearest tenth.

Answer

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: [ \begin{align*} y&=7+11.472\times0.854-\frac{1}{2}\times32\times(0.854)^2\ &=7 + 9.803-11.667\ &=5.136 \end{align*} ]

Answer:

(5.1) feet