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)
This is how I would solve it:
1st # = x 4(x+1)=3x+18
2nd # = x+1 4x+4=3x+18
4x=3x+14
x=14
To check: 15x4=60; 60-42=18. <u>1st # = 14; 2nd # = 15.</u>
