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Get Full Access to University Physics - 13 Edition - Chapter 7 - Problem 5e
Get Full Access to University Physics - 13 Edition - Chapter 7 - Problem 5e

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# A baseball is thrown from the roof of a 22.0-m-tall ISBN: 9780321675460 31

## Solution for problem 5E Chapter 7

University Physics | 13th Edition

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Problem 5E

A baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m/s and directed at an angle of 53.1o above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1o below the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?

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Solution 5E The height from where the ball was thrown is 22 m. Initial velocity was, V 0 12 m/s. Angle is, = 53.1 .0 a) Initially the potential energy of the ball was, P.E initialmgh = m × 9.8 × 22 = 215.6 m J . To get the initial velocity which has two components, we need to use a little bit of trigonometric deductions. 0 V =x12 cos 53.1 = 7.205 V =y12 sin 53.1 = 9.6. The x component remains constant through the whole motion, so we need to consider the y component only. Because the x component will get cancelled as it is same everywhere. Initially the kinetic energy was, 2 2 2 K.E initial1/2 mV y = 1/2 × m × V y = 0.5 × m × 9.6 = 40 m. The final potential energy would be, P.E final mgh = 9.8 × m × h. Final kinetic energy is zero at the highest point where the ball reached. K.E final 0. We know from the conservation of mechanical energy that, P.E initialK.E initial P.E final+ K.E final 215.6 m + 72 m = 9.8 × m × h + 0 215.6 + 72 = 9.8 h h = 287.6 / 9.8 = 29.34 meters Now, the potential energy at the highest point is , P.E = mgh = m × 9.8 × 29.34 = 287.6 m J initial The final potential energy would be zero. Initial kinetic energy was zero, because of the zero speed at the top. Final kinetic energy will be, K.E = 1/2 m v 2 final From conservation of mechanical energy, 2 287.6 m = 1/2 m v 2 v = 2 × 287.6 = 575.2 v = 575.2 = 24 m/s b) If the ball would have been thrown at an angle of 53.1 below the horizontal, then also the final velocity would be 24 m/s. It is independent of the angle and depends only on the conservation of mechanical energy. c) The final speed will be reduced because of the air resistance. We know that friction is a nonconservative force and our system will become a dissipative system as the total mechanical energy would not be conserved. Because of air friction, the speed will be reduced. None of the situation would give the higher speed.

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##### ISBN: 9780321675460

The answer to “A baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m/s and directed at an angle of 53.1o above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1o below the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?” is broken down into a number of easy to follow steps, and 89 words. This textbook survival guide was created for the textbook: University Physics, edition: 13. University Physics was written by and is associated to the ISBN: 9780321675460. This full solution covers the following key subjects: air, angle, velocity, speed, resistance. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. The full step-by-step solution to problem: 5E from chapter: 7 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 5E from 7 chapter was answered, more than 809 students have viewed the full step-by-step answer.

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