Solve.The distance an object travels after accelerating from an initial velocity is the product of velocity and the time added to one-half of the product of the acceleration and the square of the time. Write an expression for the height h of an object t seconds after being fired straight up into the air with an initial velocity of 288 m/sec. Then write the expression in factored form.[Note: The acceleration of an object fired vertically upwards is -9.8. m/s2]
A. 288t - 4.9t2 = t(288 - 4.9t)
B. 288t - 4.9t2 = 4.9t(288 - t)
C. 288t - 9.8t2 = t2(288 - 9.8t)
D. 288t - 9.8t2 = t(288 - 9.8t)
Answer: A
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