How much force is needed to accelerate a space shuttle 15 m/s^2 with a mass of 2,030,000 kg? I will give you brainly if correct
1 answer:
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Answer:
a) 49 ft
b) 66 ft
c) 4 seconds
d) [0, 4] seconds
Step-by-step explanation:
a) Evaluate the function for t=3:
h(3) = -16·3² +63·3 +4 = (-16·3 +63)·3 +4 = 15·3 +4
h(3) = 49
The height of the ball is 49 feet after 3 seconds.
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b) The maximum height of the ball will be found where t=-b/(2a) = -63/-32 = 1.96875.
h(1.96875) = (-16·(63/32) +63)·(63/32) +4 = 63²/64 +4 = 66.015625
The maximum height of the ball is approximately 66 feet.
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c) The ball will hit the ground when its height is zero.
-16t² +63t +4 = 0
Using the quadratic formula, we find the solution to be ...
t = (-63 - √(63² -4(-16)(4)))/(2·(-16)) = (-63 -√4225)/-32 = -128/-32 = 4
The ball will hit the ground after 4 seconds.
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d) The function is only useful for the time period between when the ball is thrown and when it lands, t = 0 to t = 4 seconds.
The domain of t in the interval 0 to 4 seconds makes sense for this function.
