hf=Hi+vi*t-1/2 9.8 t^2
Hi=125
vi=5.2
hf=0
solve for t. Notice it is a quadratic equation, use the quadratic equation.
A helicopter is ascending vertically with a speed of 5.20 m/s. At a height of 125 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground? [Hint: The package's initial speed equals the helicopter's.]
4 answers
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Given:
Vi=5.20m/s
h=125 m
g=9.8m/s²
Formula:
t=h-Vi ÷(over)½g
t=125m - 5.20m/s ÷(over) ½(9.8m/s²)
t=119.8s÷½(9.8s²)
t=119.8s÷4.9s²
t=24.45s
Explanation:
How did I get the answer 119.8s?
I subtracted 125 m and 5.20 m/s than cancelled out the meter.
How did I get the answer 4.9s²?
I divided 9.8m/s² by 2 simply getting it from the ½.
How I ended up with the answer 24.45s?
I simply cancelled out the second from the ² power of 4.9s² and cancelled out the s from 119.8s to have the answer.
Vi=5.20m/s
h=125 m
g=9.8m/s²
Formula:
t=h-Vi ÷(over)½g
t=125m - 5.20m/s ÷(over) ½(9.8m/s²)
t=119.8s÷½(9.8s²)
t=119.8s÷4.9s²
t=24.45s
Explanation:
How did I get the answer 119.8s?
I subtracted 125 m and 5.20 m/s than cancelled out the meter.
How did I get the answer 4.9s²?
I divided 9.8m/s² by 2 simply getting it from the ½.
How I ended up with the answer 24.45s?
I simply cancelled out the second from the ² power of 4.9s² and cancelled out the s from 119.8s to have the answer.
1 answer