One object is fired at an angle of ten degrees above the horizontal. A second object is fired at an angle of ten degrees below the horizontal, and a third is fired in an exactly horizontal direction. All three are fired at the same time, but with different (and nonzero) speeds. Neglect air resistance and assume that the objects are fired over a perfectly level plain. Which of the objects strikes the ground first?A.The one fired at an angle of ten degrees above the horizontal.B.The one fired at an angle of ten degrees below the horizontal.C.The one fired exactly horizontally.D.They will all strike the ground at the same time.E. You cannot tell from the information given.F. None of the answers above is correct.

We can explain the situation by using the principles of projectile motion in which an object is launched in free air with some angle measured from the horizontal. The only force acting to modify the velocity of the object is the force of gravity. It only acts on the vertical axis. When the angle is positive, the object goes up the launch level, stops in mid-air, and finally goes towards the Earth's surface. If the angle is zero, no upward motion is achieved, the object starts falling immediately from zero speed and increating its speed until hitting the ground. Finally, if the angle is negative, the object also starts falling immediately but with some initial speed pointing downwards. This object must reach ground level first, regardless of the initial value of the speed.

This can be also proven by handling the formula of the height (y)

[tex]\displaystyle y=y_o+v_{oy}t-\frac{gt^2}{2}[/tex]

Where [tex]y_o[/tex] is the initial height of launching, [tex]v_{oy}[/tex] is the initial speed in the vertical axis, and t is the time elapsed since the launching started.

The value of [tex]v_{oy}[/tex] can be computed as

[tex]v_{oy}=v_o\ sin\theta[/tex]

Where [tex]\theta[/tex] is the angle with the horizontal. If it is positive, [tex]v_{oy}[/tex] is positive too. If [tex]\theta[/tex] is zero, [tex]v_{oy}[/tex] is zero, and if [tex]\theta[/tex] is negative, [tex]v_{oy}[/tex] is also negative. It we rearrange the formula of y

[tex]\displaystyle y-y_o+\frac{gt^2}{2}=v_{oy}t[/tex]

The left side is the same for the three objects no matter their initial speed or angle, the difference of heights at any time t depends on the right-hand term. The object with less height will always be the one whose angle is negative, the second less height corresponds to the object launched with angle zero. The last to hit the ground will be the object with a positive angle. These results are independent of the value of [tex]v_o[/tex],

Answer:

B. The one fired at an angle of ten degrees below the horizontal