If a ball is thrown into the air with a velocity of 42 ft/s?
2016-07-03 4:43 pm
If a ball is thrown into the air with a velocity of 42 ft/s, its height (in feet) after t seconds is given by y = 42t − 16t2. Find the velocity when t = 1.
回答 (3)
2016-07-03 4:59 pm
y = 42t − 16t²
velocity, v(t) = dy/dt = (d/dt)(42t − 16t²) = 42 - 16t
When t = 1 :
velocity, v(1) = 42 - 16(1) = 26 (ft/s)
velocity, v(t) = dy/dt = (d/dt)(42t − 16t²) = 42 - 16t
When t = 1 :
velocity, v(1) = 42 - 16(1) = 26 (ft/s)
2016-07-03 4:51 pm
Hmm. Lets see.... 42 x 1 - 16 x 1 x 1 = 42 - 16 = .....
Maybe you can do the last step?
Maybe you can do the last step?
2016-07-03 4:52 pm
y = 42 * t – 16 * t^2
Let t be1
y = 42 * 1 – 16 * 1^2 = 26 ft
y = vi * t – ½ * g * t^2
½ * g = 16
g = 32 ft/s^2
This is value of g in this unit.
Let t be1
y = 42 * 1 – 16 * 1^2 = 26 ft
y = vi * t – ½ * g * t^2
½ * g = 16
g = 32 ft/s^2
This is value of g in this unit.
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