Where Will the Falling Suitcase Land? - Projectile Motion Homework Question

[alex2256]

Homework Statement

OK, I'm posting to clarify if my answer is correct.

1. An airplane is flying with a velocity of 90.0 m/s at an angle of 23.0m\s above the horizontal. When the plane is a distance 115 m directly above an observer that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the observer will the suitcase land? You can ignore air resistance.

Homework Equations

y = ut + 1/2at^2
x = ut

The Attempt at a Solution

OK, I divided the x and y components into a table, then solved for time and got 4.845 seconds. I then plugged it into x = ut and got 436.05m, then constructed a triangle and did the Pythagorean Theorem and got 450.9m

Could someone please check this for me, since I don't have the answers to the question?

Thanks.

 

[rl.bhat]

To find time of flight t, use
y = - u*sinθ*t + 1/2*g*t^2.
- u*sinθ is the vertical component of the velocity of projection.
Then the range = u*cosθ*t

 

[nlightner]

Your approach seems correct. To verify your answer, you can also use the equations for projectile motion:

y = y0 + v0y*t + 1/2 * a * t^2
x = x0 + v0x * t

Where:
y0 = initial height (115m)
v0y = initial vertical velocity (calculated using 90m/s and 23 degrees)
a = acceleration due to gravity (-9.8 m/s^2)
x0 = initial horizontal position (0m)
v0x = initial horizontal velocity (calculated using 90m/s and 23 degrees)

Solving for t in the first equation, we get t = 4.845 seconds, which is the same as what you calculated. Plugging this value of t into the second equation, we get x = 436.05m, which is also the same as what you calculated.

So, your answer of 450.9m using the Pythagorean Theorem is correct. Good job!