y=gt²/2
t=sqrt(2y/g) =sqrt(2•920/9.8)=13.7 s
A cargo plane is moving with a horizontal velocity of
vx = +225 m/s
at a height of
y = 920 m
above level ground as shown in the figure below when it releases a package. Ignoring air resistance, how much time will it take the package to reach the ground? (Express your answer to the nearest tenth of a second.)
2 answers
For this question you're going to need a kinematic equation. Because you're solving for time, I would use d=Voy*t+0.5*a*t^2
The acceleration is 9.8 m/s^2 (making "down" positive) due to gravity.
There is no initial velocity (V) in the y direction.
You do not need to use the fact that Vox is 225 m/s.
So if d=Voy*t+0.5*a*t^2 first substitute the values that you know. You get 920=0.5*(9.8)*t^2
solve for t. You should get something like 13.7 seconds.
Hope that helps!
The acceleration is 9.8 m/s^2 (making "down" positive) due to gravity.
There is no initial velocity (V) in the y direction.
You do not need to use the fact that Vox is 225 m/s.
So if d=Voy*t+0.5*a*t^2 first substitute the values that you know. You get 920=0.5*(9.8)*t^2
solve for t. You should get something like 13.7 seconds.
Hope that helps!
2 answers