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

think its the third option

unbalanced force

explanation: