An airplane is flying with a velocity of 95.0m/s at an angle of 22.0 degreesabove the horizontal. When the plane is a distance 116m m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.
How far from the dog will the suitcase land?
ignore air resistance

Question

An airplane is flying with a velocity of 95.0m/s at an angle of 22.0 degreesabove the horizontal. When the plane is a distance 116m m directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment.
How far from the dog will the suitcase land? 
ignore air resistance

anonymous · Answer

Ok, let me see : 

First you need to decompose the velocity : $ \Large V = 95$ m/s 

$\Large V_x = 95\times cos(22) = 88.08$ m/s 

$\Large V_y = 95\times sin(22) = 35.59$ m/s 

The suitcase first will go up, and then it will go down, the suitcase is $\Large 116$ over the ground, now, let's work with the vertical velocity, let's find the maximum altitude, the time the suitcase takes, and then, let's find the total flying time : 

$\Large V_y = 35.59m/s $

maximum altitude : 

$\Large Vf^2 = V_y^2 - 2\times g \times h$ 

considering $\Large g = 9.8 m/s^2$ 

When the suitcase reaches its maximum altitude, $\Large Vf = 0 m/s$ 

$\Large 0^2 = 35.59^2 - 2\times9.8\times h$ 

$\Large h = 64.62$ meters 

$\Large Vf = V_y - g\times t$ 

$\Large 0 = 35.59 - 9.8\times t$ 

$\Large t = 3.63$ seconds 

Now, let's find the time, the suitcase took to go down : 

total altitude $\Large = 116 + 64.62 = 180.62$ meters 

$\Large 180.62 = 1/2\times9.8\times t'^2$ 

$\Large t' = 6.07$ seconds 

Total flying time $\Large = T = 6.07 + 3.63 = 9.7$ seconds 

Now, we can find how far the suitcase landed : 

$\Large X = V_x\times T$ 

$\Large X = 88.08\times9.7 = 854.376$ meters 

Hope that helps

anonymous · Answer

thank you so much! @junghyunran