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 40
ft per sec. The height s of the ball in feet is given by the equation s equals negative 2.7 t squared plus 40 t plus 6.5s=−2.7t2+40t+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 12 ft above the moon's surface?
1 answer:
Response:
0.14 s
Detailed breakdown:
s = -2.7 t² + 40t + 6.5
Set s = 12
12 = -2.7t² + 40t + 6.5 Rearranging yields
-2.7t² + 40t + 6.5 - 12 = 0
-2.7t² + 40t - 5.5 = 0
Utilize the quadratic formula
a = -2.7; b = 40; c = -5.5
x = 7.41 ± 7.27
x₁ = 0.14; x₂ = 14.68
The graph indicates roots at x₁ = 0.134 and x₂ = 14.68.
The surface of the Moon stands at -12 ft. Thus, the ball will reach a height of 12 ft above the Moon’s surface (crossing the x-axis) at 0.14 s.
The second root indicates when the ball is again 12 ft above the lunar surface as it descends.
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Answer:
8
Step-by-step explanation:
The task is to evaluate:
0.1m + 8 - 12n
When
By substituting these values into the expression, we have: