Explanation Given: Rock passes the observer exactly at t = 4 sec Acceleration due to gravity on moon (g) = 5.5 ft/s 2 Concept used: Negative value of acceleration due to gravity is used when stone is thrown upward or against the gravity. Drawing a diagram of the given problem according to given statement: Note (Some observation from the above figure ): At time t = 0 velocity of the rock = 0 But,just after the time t = 0 there is some velocity in the stone so, we assumed that velocity as initial velocity of the stone. As stone is taking exactly 4 sec. to pass the observer after throwing so it reaches the highest peak at time (t) = 2 sec. Because it will take equal time from initial position to reach highest point and from highest point to pass the observer. Velocity of the stone at highest point exatly at time 2 sec. is zero. Calculation: Now, calculating the initial velocity (u) of the stone just after the throwing from the cliff: v = u + a t (formula) where; v → velocity of the stone exactly time t = 2 sec ( v = 0 ) . u → initial velocity of the stone a → acceleration due to gravity in upward direction ( a = − 5.5 ft/s 2 ) put all the values in the above formula v = u + a t 0 = u + ( − 5.5 ) × 2 u = 11 ft/s 2 Hence, the initial velocity of the stone is 11 ft/s 2

7th Edition

ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Differential Equations

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Chapter 3.4, Problem 44PS

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To find : the initial velocity of the rock and height of the cliff from the moon surface.

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Chapter 3.4, Problem 43PS

Chapter 3.4, Problem 45PS

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Chapter 3 Solutions

Calculus

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the initial velocity of the rock and height of the cliff from the moon surface.

Solution Summary: The author calculates the initial velocity and height of the cliff from the moon surface.

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To find : the initial velocity of the rock and height of the cliff from the moon surface.

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Chapter 3 Solutions