If a ball is thrown directly upward with a velocity of 30 ft/s, its height (in feet) after t seconds is given by y = 30t − 16t^{2}. What is the maximum height attained by the ball? (Round your answer to the nearest whole number.)

Answer:

the maximum height is attained at time t such that the speed is zero, namely:

0 = 30 - 32.2 *t, since gravity g = 32.2 feet/sec^2

so:

t = 30/32.2 = 0.93 seconds,

so y_max = 30 * 0.93- 16* 0.93^2 = 27.9 - 13.84 = 14.06 feet.

SO the maximum height is y_max = 14.06 feet

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