Asked by Emily
An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 50 ft per sec. The height s of the ball in feet is given by the equation s=-2.7t^2+50t+6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b.
A.) After how many seconds is the ball 10ft above the moon's surface?
B.) how many seconds will it take for the ball to hit the moon's surface?
A.) After how many seconds is the ball 10ft above the moon's surface?
B.) how many seconds will it take for the ball to hit the moon's surface?
Answers
Answered by
mathhelper
a)
10 = -2.7t^2 + 50t + 6.5
2.7t^2 - 50t + 3.5 = 0
solve for t , using the quadratic formula, remember you have 2
positive answers,
both are valid, one is on the way up, the other on the way down
b) solve
-2.7t^2 + 50t + 6.5 = 0 , again using the quadratic formula
10 = -2.7t^2 + 50t + 6.5
2.7t^2 - 50t + 3.5 = 0
solve for t , using the quadratic formula, remember you have 2
positive answers,
both are valid, one is on the way up, the other on the way down
b) solve
-2.7t^2 + 50t + 6.5 = 0 , again using the quadratic formula
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