An airplane is flying with a velocity of 100 m/s at an angle of 25° above the horizontal. when the plane is 114 m directly above a dog that is standing on horizontal ground, a suitcase drops out of the luggage compartment. how far from the dog will the suitcase land? ignore air resistance

The suitcase will land 976.447m from the dog.

Explanation:

The velocity in its component in the X and Y axis is decomposed:

Vx= 100m/s × cos(25°)= 90.63m/s

Vy= 100m/s × sen(25°)= 42.26m/s

Time it takes for the suitcase to reach maximum height, the final speed on the axis and at the point of maximum height is zero whereby:

VhmaxY= Voy- 9.81(m/s^2) × t ⇒ t= (42.26 m/s) / (9.81(m/s^2)) = 4.308s

The space traveled on the axis and from the moment the suitcase is thrown until it reaches its maximum height will be:

Dyhmax= Voy × t - (1/2) × 9.81(m/s^2) × (t^2) =

= 42.26m/s × 4.308s - 4.9 (m/s^2)  × (4.308s)^2 =

=182.056m - 90.938m= 91.118m

The time from the maximum height to touching the ground is:

Dtotal y= 114m + 91.118m = (1/2) × 9.81(m/s^2) × (t^2) =

= 205.118m = 4.9 (m/s^2) × (t^2) ⇒ t= (41.818 s^2) ^ (1/2)= 6.466s

The total time of the bag in its rise and fall will be:

t= 4.308s + 6.466s = 10.774s

With this time and the initial velocity at x which is constant I can obtain the distance traveled by the suitcase on the x-axis:

Dx= 90.63 (m/s) × 10.774s = 976.447m

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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