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=−2.7t^2+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 22 ft above the moon's surface?
After seconds the ball will be 22 ft above the moon's surface.
b. How many seconds will it take for the ball to hit the moon's surface?
It will take _ seconds for the ball to hit the moon's surface.

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akinyemisakiru akinyemisakiru · Answer 1 · 2020-01-31T19:32:38+01:00

Answer:

a) 14.59s or 0.23s

b) 29.63s

Step-by-step explanation:

Let the height of the astronaut H = 6.5ft

Assuming the astronaut released the ball immediately at his height,

a) when the height is 22ft above the moon surface,

S = 22 - 6.5 =−2.7t^2+40t+6.5

2.7t^2-40t+ 9= 0

Using quadratic formular,

t = 14.59s or 0.23s

Therefore,

After 0.23s the ball will be at 22 ft above the moon surface immediately it is released by the astronaut

After 14.59s the ball will be at 22ft above the moon surface after making return from the maximum height

b) To calculate the time the ball will hit the moon surface, we need to obtain the maximum height.

Differentiate, s=−2.7t^2+40t+6.5

ds/dt = -5.4t + 40

At maximum height, ds/dt = 0

t = 40/5.4 = 7.41 s

Input t = 7.41 to obtain the maximum S,

s=−2.7t^2+40t+6.5 = -2.7(7.41)² +40(7.41) + 6.5

S = 154.65ft

Therefore, the time to travel the astronaut height 6.5 ft

S= 6.5 ft =−2.7t^2+40t+6.5

2.7t^2- 40t = 0

t = 40/2.7 = 14.81s

Therefore, it will take t= 14.81 +7.41+7.41 = 29.63s for the ball to hit the moon surface

t = 29.63s