phys 1111 - clickers - 2d motion\nfrom the same height and at the same time, one ball is dropped and another…

phys 1111 - clickers - 2d motion\nfrom the same height and at the same time, one ball is dropped and another is fired horizontally.\nwhich one will hit the ground first?\na. the “dropped” ball\nb. the “fired” ball\nc. they both hit at the same time\nd. it depends on how hard the ball was fired\ne. it depends on the initial height

phys 1111 - clickers - 2d motion\nfrom the same height and at the same time, one ball is dropped and another is fired horizontally.\nwhich one will hit the ground first?\na. the “dropped” ball\nb. the “fired” ball\nc. they both hit at the same time\nd. it depends on how hard the ball was fired\ne. it depends on the initial height

Answer

Explanation:

Step1: Analyze vertical motion

The vertical - motion of both balls is a free - fall motion. The initial vertical velocity of the dropped ball is $v_{0y1}=0$, and the initial vertical velocity of the horizontally - fired ball is also $v_{0y2}=0$. The vertical displacement for both balls is $y - y_0=-h$ (taking downwards as negative), and the acceleration due to gravity is $a = - g$.

Step2: Use the kinematic equation

The kinematic equation for vertical displacement is $y - y_0=v_{0y}t+\frac{1}{2}at^{2}$. For both balls, $y - y_0=-h$, $v_{0y} = 0$, and $a=-g$. Substituting these values into the equation gives $-h = 0\times t-\frac{1}{2}gt^{2}$, or $t=\sqrt{\frac{2h}{g}}$. Since the height $h$ and the acceleration due to gravity $g$ are the same for both balls, the time it takes for them to reach the ground is the same.

Answer:

C. They both hit at the same time