S = ut + (1/2)a(t²) Subtract ut from both sides
(1/2)a(t²) = S - ut Multiply both sides by 2
a(t²) = 2s - 2ut Divide both sides by t²
a= 2s/t² - 2u/t
a= (2S - 2ut)/t²
Answer is C) but there should be parentheses around the term (2S-2ut)
The answer to this mathematical question is 3
Answer:
Step-by-step explanation:
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Collecting like terms and expanding the brackets
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Answer:
t₁ = 2,75 sec
Step-by-step explanation:
The ball will get h maximum when dh/dt ( its vertical component of the speed is equal to 0. At this moment the ball has flown half of the total flight time
Then
h(t) = - 4t * ( 4*t - 11 )
h(t) = - 16*t² + 44*t
Taking derivatives on both sides of the equation we get:
dh/dt = - 32*t + 44
dh/dt = 0 ⇒ -32*t + 44 = 0
t = 44/32 ⇒ t = 1,375 s
So twice this time
2*t = 2 * 1.375
Total time t₁ = 2,75 sec
Since we have the length and width of the window, we can multiply the two values.
Convert both fractions to improper fractions
(26/3)*(23/4)
=598/12 ft squared
= This simplifies to 49 5/6 ft. squared
