- Mr. Muscedere hits a golf ball with an initial velocity of 25m/s at an angle of 30.0 degrees above the horizontal. The tee is 14m above the green. • What is the maximum height of the ball? • What is the velocity of the ball when it lands?

- Mr. Muscedere hits a golf ball with an initial velocity of 25m/s at an angle of 30.0 degrees above the horizontal. The tee is 14m above the green. • What is the maximum height of the ball? • What is the velocity of the ball when it lands?

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Description: Please answer question and just send me the paper solutions ASAP please faster and use these equation • Mr. Muscedere hits a golf ball with an initial velocity of 25m/s at an angleof 30.0 degrees above the horizontal. The tee is 14m above the green.•

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter4: Two-and-three Dimensional Motion

Section: Chapter Questions

Problem 51PQ

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Problem For a projectile lunched with an initial velocity of v0 at an angle of θ (between 0 and 90o) , a) derive the general expression for maximum height hmax and the horizontal range R. b) For what value of θ gives the highest maximum height? Solution The components of v0 are expressed as follows: vinitial-x = v0cos(θ) vinitial-y = v0sin(θ) a) Let us first find the time it takes for the projectile to reach the maximum height. Using: vfinal-y = vinitial-y + ayt since the y-axis velocity of the projectile at the maximum height is vfinal-y = ? Then, ? = vinitial-y + ayt Substituting the expression of vinitial-y and ay = -g, results to the following: ? = ? - ?t Thus, the time to reach the maximum height is tmax-height = ?/? We will use this time to the equation yfinal - yinitial = vinitial-yt + (1/2)ayt2 if we use the time taken to reach the maximum height, therefore, the displacement will yield the maximum height, so hmax = vinitial-yt + (1/2)ayt2 substituting, the vinitial-y…