An airplane is flying with a velocity of \(90m/s\) at an angle of \(23^\circ\) above the horizontal. When the plane is \(114m\) 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.

**Solution:**

The suitcase moves in projectile motion. The initial velocity of the suitcase is equal to the velocity of the airplane.

To find the time, it takes to reach the ground

the y-component of the velocity $$ v_{0y}=v_0sin23^\circ$$

The acceleration $$ a_y=-9.8m/s^2$$ since +y is taken to be upward.

Now, the vertical distance from the plane to the dog s=114m.

Putting these values in the equation of motion to find the time t, we have

$$ 114=90sin23^\circ+\frac{1}{2}\times(-9.8)\times t^2$$

This gives, $$ t=9.60s$$

TThe distance the suitcase travels horizontally is $$v_{0x}t=(v_0 cos23^\circ)t=795m$$

You have use formula s=ut+at^2 But you have not multiply usinA with t