Projectile motion airplane speed problem

[dethlok777]

An airplane with a speed of 97.5 m/s is climbing upward at an angel of 50.0 degrees with respect to the horizontal. When the plane's altitude is 732m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

I started this problem by figuring out the velocity components for both the horizontal motion and the vertical motion of the package.
initial velocity of x=the inital velocity*cos(50.0 degrees)=62.7 m/s
initial velocity of y= the inital velocity *sin (50.0 degrees)=74.7 m/s
Then I thought I should find the time it takes for the package to hit the ground. using the equation y=initial velocity component y*t + 1/2(acceleration component of y*t*t)-->-732m=(74.7m/s)t + 1/2(-9.80m/(s*s))t*t = 22s
My question: Is this the direction I should be following to solve this problem or am I totally hosed up on this? Don't want the answer to the overall question but a hint to help me figure this out.
ps sorry about the t*t notation I could not figure out how to do superscripts or subscripts here.

 

[Q_Goest]

Yes, you've got a good start on this problem, and resolving the velocities into X and Y is the right way to go. Now you have to determine the Y velocity at the time it hits the ground. With that, you have both the X and Y velocity just before impact, assuming the X velocity hasn't changed and neither velocity is a function of wind resistance. From that you can find the angle it is going upon impact, just as you resolved the X and Y from the initial angle.

 

[quicknote]

Yes, you are on the right track! To solve this problem, you will need to use the equations of motion for projectile motion. The first step, as you have correctly done, is to find the initial velocity components in the horizontal and vertical directions. From there, you can use the equations of motion to find the time it takes for the package to hit the ground and the distance it travels horizontally.

One thing to keep in mind is that the vertical acceleration in this case is not just -9.8 m/s^2, as it would be for an object in free fall. The plane is still moving upward at an angle, so the acceleration is actually a combination of gravity and the upward motion of the plane. This means that the vertical acceleration will be less than -9.8 m/s^2, and you will need to use the angle of the plane's motion to calculate the correct value.

Once you have the time and distance, you can use trigonometry to find the angle of the velocity vector just before impact. Remember that the horizontal and vertical components of the velocity will change as the package falls, but the overall velocity magnitude will remain constant.

Overall, you are on the right track with your approach, and it will just take some careful calculations and equations to solve the problem. Good luck!