Question

question 2
two students are standing on a fire escape, one twice as high as the other. simultaneously, each drops a ball. if the first ball strikes the ground at time dt, when will the second ball strike the ground? (disregard air resistance. assume a = -g=-9.81 m/s²)
o dt = √2dt
o dt = 1/√2dt
o dt = 4dt
o dt = 2dt
question 3
which would hit the ground first if dropped from the same height in a vacuum - a feather or a metal bolt?
o they would hit the ground at the same time.
o the feather
o the metal bolt.
o they would be suspended in a vacuum.

Explanation:

Question 2

Step1: Use free - fall formula

The free - fall formula is $h = v_0t+\frac{1}{2}at^2$. Since the balls are dropped ($v_0 = 0$), the formula simplifies to $h=\frac{1}{2}gt^2$, or $t=\sqrt{\frac{2h}{g}}$. Let the height of the first student be $h_1$ and the time it takes for the first ball to fall be $t_1=\sqrt{\frac{2h_1}{g}}$. The height of the second student is $h_2 = 2h_1$, and the time it takes for the second ball to fall is $t_2=\sqrt{\frac{2h_2}{g}}=\sqrt{\frac{2\times(2h_1)}{g}}=\sqrt{2}\times\sqrt{\frac{2h_1}{g}}$. Since $t_1=\sqrt{\frac{2h_1}{g}}$, then $t_2=\sqrt{2}t_1$.

In a vacuum, there is no air resistance. All objects fall with the same acceleration due to gravity ($g$). Using the free - fall equation $h = v_0t+\frac{1}{2}at^2$ (with $v_0 = 0$), the time of fall $t=\sqrt{\frac{2h}{g}}$ depends only on the height $h$ and the acceleration due to gravity $g$. Since the height is the same for both the feather and the metal bolt, they will hit the ground at the same time.

Answer:

$\sqrt{2}t_1$

Question 3