The formula for the vertical height, h, attained with a vertical take off speed of u is
u² - 2gh = 0
where
g = 9.8 m/s², acceleration due to gravity.
Because h = 1.29 m, therefore
u² = 2*9.8*1.39 = 27.244
u = 5.2196 m/s
Also, the time of flight, t, required when the vertical take-off speed is u is given by
ut - 0.5gt² = 0.
Therefore
5.2196t - 0.5*9.8*t² = 0
5.2196t - 4.9t² = 0
t(5.2196 - 4.9t) = 0
t = 0, or t = 5.2196/4.9 = 1.0652 s
t = 0 corresponds to take-off.
t = 1.0652 s corresponds to landing.
The hang time is 1.0652 s.
Answer:
Take-off speed = 5.22 m/s (nearest hundredth)
Hang time = 1.065 s (nearest thousandth)
Answer:
Step-by-step explanation:
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For this case we must express the following expression algebraically:
<em>"The quotient of b and 2 minus 4 is at least 26"</em>
So we have to:
The quotient of b and 2 minus 4, is represented as:
We have different signs subtracted and the sign of the major is placed:
Thus, the expression is written as:
ANswer:
3/4x + 5 - 1/2x = 3
3/4x - 1/2x = 3 - 5
3/4x - 2/4x = -2
1/4x = -2
x = -2 / (1/4)
x = -2 * 4/1
x = -8 <==
