htt rationalreasoning.net/assessment/showtest php?action skip&ito-6; 61 Fall 201

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htt rationalreasoning.net/assessment/showtest php?action skip&ito-6; 61 Fall 2017 Due in 3 hours, 11 minutes. Due Mon 10/16/2017 11-59 A penny is thrown from the top of a 48.8-meter building and hits the ground 2.27 seconds after it was thrown. The penny reached its maximsum beight above the ground 0.51 seconds after it was thrown. a. Define a quadratic function, h, that expresses the height of the penny above the ground (measured in meters) as a function of the number of seconds elapsed since the penny was thrown, t. Preview b. What is the maximum height of the penny above the ground? Points possible: 10 Unlimited attempts. Post this question to forum Submit 850 PM 10/16/2017

Explanation / Answer

as the maximum height is attained at t = 0.51 s after throwing. Velocity must be zero at this point,

let initial vertical velocity = u

then, u = 0.51g

also h = 48.8 + ut - 0.5gt2

=> h = 48.8 + 0.51gt - 0.5gt2

=> h = 48.8 + 4.998t - 4.9t2

now at t = 0.51 s, h = 48.8 + 4.998 x 0.51 - 4.9 x 0.512 = 50.07449 m

Note: there are two time values given which both give contradictory results. Though one is sufficient t calculate the required problem. Here I used the time for max height i.e 0.51 s

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