Explanation Given data: Horizontal distance, the brick travelled = 44.5 m Neglect the effects caused by air resistance. The brick landed at the same height it was launched. Formula used: s = u t + 1 2 a t 2 where s is the displacement, u is the initial velocity, t is the time and a is the acceleration. v = u + a t where v is the displacement, u is the initial velocity, t is the time and a is the acceleration. Calculation: Consider the projectile motion with the initial velocity being u and the angle with the horizontal being θ as shown in the figure. Applying the equation s = u t + 1 2 a t 2 for the vertical direction for the whole motion, s = u t + 1 2 a t 2 0 = u sin θ t − 1 2 g t 2 t ( 1 2 g t − u sin θ ) = 0 As t ≠ 0 1 2 g t − u sin θ = 0 t = 2 u sin θ g Applying the equation s = u t + 1 2 a t 2 for the horizontal direction for the whole motion s = u t + 1 2 a t 2 r = u cos θ t + 0 r = u cos θ t Combining equations t = 2 u sin θ g and r = u cos θ t r = u cos θ × 2 u sin θ g As r = 44.5 m and g = 9.81 m

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ISBN: 9781429281843

Author: Tipler

Publisher: MAC HIGHER

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Chapter 3, Problem 37P

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To Find: The speed of the brick at the highest point of its flight.

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Chapter 3, Problem 36P

Chapter 3, Problem 38P

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The speed of the brick at the highest point of its flight.

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