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25 November, 20:56

On Mars, the gravity acting on an object is less than that on Earth. On Earth, a golf ball hit with an initial upward velocity of 26 meters per second will hit the ground in about 5.4 seconds. The height h of an object on Mars that leaves the ground with an initial velocity of 26 meters per second is given by the equation h = - 1.9t 2 + 26t. How much longer will it take for the golf ball hit on Mars to reach the ground? Round your answer to the nearest tenth.

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  1. 25 November, 21:18
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    The answer to the question is

    It take for the golf ball hit on Mars to reach the ground 8.284 s longer.

    Step-by-step explanation:

    To Solve the question, we note that the equation of motion is given to us as

    h = - 1.9·t² + 26·t

    Therefore the height

    We compare the above equation with the equation of of motion

    S = ut - 0.5*g*t² where u = initial velocity = 26 m/s and - 0.5*g = - 1.9

    Therefore the acceleration due to gravity om mars = - 1.9/-0.5 = 3.8 m/s²

    We now have the time taken to maximum height given by

    v = u - g·t where v = 0 at maximum height

    We therefore have u = g·t → 26 = 3.8*t or t = 26/3.8 = 6.84 seconds

    The time taken for ball to hit the ground is 2 * the time to reach maximum height = 2 * 6.84 = 13.684 s

    The time longer it takes for the ball on Mars to complete the path up to maximum height and down again is 13.684 s - 5.4 s = 8.284 s.
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