Answer:
[tex] t =\sqrt{\frac{2*(-100m)}{-9.8 m/s^2}}= 4.518 s[/tex]
And now we can find the final distance on the x axis using the formula:
[tex] D = V_x t[/tex]
The velocity on x not changes and is the same plane spped [tex] V_x = 150 m/s[/tex], if we replace we got:
[tex] X= 150 m/s * 4.518 s= 677.63 m[/tex]
Explanation:
For this case we have a illustration for the problem in the figure attached.
We need to find how far short of the target should be the plane, we have the following info given:
[tex] v_{ix}= 150 m/s[/tex]
[tex] v_{iy}=0 m/s[/tex]
[tex] y_{f} =-100m[/tex]
[tex] y_i =0m[/tex]
[tex] x_i = 0m[/tex]
How far short of the target should it drop the package?
First we can find the total time in order to reach the groung using the following kinematic formula:
[tex] y_f = y_i + v_{iy} +\frac{1}{2}gt^2[/tex]
And replacing we have:
[tex] -100 m = 0 +0 -\frac{1}{2} (9.8 m/s^2) t^2[/tex]
And solving for t we got:
[tex] t =\sqrt{\frac{2*(-100m)}{-9.8 m/s^2}}= 4.518 s[/tex]
And now we can find the final distance on the x axis using the formula:
[tex] D = V_x t[/tex]
The velocity on x not changes and is the same plane spped [tex] V_x = 150 m/s[/tex], if we replace we got:
[tex] X= 150 m/s * 4.518 s= 677.63 m[/tex]
