Answer:
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Answer:
Translation would be your answer
Step-by-step explanation:
<u>Answer</u> : The demonstration is below :)
Step-by-step explanation :
<u>We use Pythagoras' </u><u>theorem </u><u>:</u>
- In the triangle ABC we have :
AB² = AC² - BC² = 15² - 9² = 144 = 12²
- In the triangle ABD we have :
DB² = AD² - AB² = 13² - 12² = 5²
cos(a) = BD/AD = 5/13
Answer:
0.293 s
Step-by-step explanation:
Using equations of motion,
y = 66.1 cm = 0.661 m
v = final velocity at maximum height = 0 m/s
g = - 9.8 m/s²
t = ?
u = initial takeoff velocity from the ground = ?
First of, we calculate the initial velocity
v² = u² + 2gy
0² = u² - 2(9.8)(0.661)
u² = 12.9556
u = 3.60 m/s
Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.
The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m
y = 0.556 m
u = 3.60 m/s
g = - 9.8 m/s²
t = ?
y = ut + (1/2)gt²
0.556 = 3.6t - 4.9t²
4.9t² - 3.6t + 0.556 = 0
Solving the quadratic equation
t = 0.514 s or 0.221 s
So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are
0.221 s, on his way to maximum height and
0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).
Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.
I believe the answer is D because if you use the largest number in that group of numbers and divide that by 6 you get 6.6666 which is less than 7
Therefore the answer is D
