A rock is thrown upward with a velocity of 21 meters per second from the top of a 43 meter high cliff, and it misses the cliff on the way back down. When will the rock be 5 meters from the water, below? Round your answer to two decimal places.

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Lane Bradley

15 October, 16:45

A rock is thrown upward with a velocity of 21 meters per second from the top of a 43 meter high cliff, and it misses the cliff on the way back down. When will the rock be 5 meters from the water, below? Round your answer to two decimal places.

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Marquise Chambers · Answer 1

5.66 seconds from when the rock was thrown. First construct an expression to represent the distance from the water to the rock. First account for the starting distance. d = 43 + ? Now add in the distance based upon the initial velocity d = 43 + 21t + ? Now add in the acceleration due to gravity d = 43 + 21t - 0.5at^2 Now substitute the known values into the formula. 5 = 43 + 21t - 0.5*9.8*t^2 Subtract 5 from both sides and perform the multiplication 0 = 38 + 21t - 4.9t^2 We now have a quadratic function with a=-4.9, b=21, c=38. Using the Quadratic formula, we can find the solutions at t = 5.66 and t=-1.37 Since the negative time doesn't make sense, the correct value will be t = 5.66 seconds.