ball is dropped at time $\mathrm{t}=0.0 \mathrm{~s}$. At $\mathrm{t}=2.0 \mathrm{~s}$, a second ball is thrown downward with speed $\mathrm{v}$. What is $\mathrm{v}$ if at $\mathrm{t}=4.0 \mathrm{~s}$, the two balls are at the same vertical position?
A) $25 \mathrm{~m} / \mathrm{s}$
B) $29 \mathrm{~m} / \mathrm{s}$
C) $49 \mathrm{~m} / \mathrm{s}$
D) $15 \mathrm{~m} / \mathrm{s}$

Question

Quizplus · Accepted Answer

The first ball is in free fall for 4 seconds, so its position can be found using the equation $y = \frac{1}{2}gt^2$, where $g = 9.8 \, \text{m/s}^2$. Substituting $t = 4$ seconds, we get $y = \frac{1}{2} \times 9.8 \times 4^2 = 78.4 \, \text{m}$. The second ball is thrown downward after 2 seconds and needs to cover the same distance in 2 seconds to be at the same position as the first ball at $t = 4$ seconds. Using the equation $y = vt + \frac{1}{2}gt^2$ for the second ball, where $t = 2$ seconds and $y = 78.4 \, \text{m}$, we can solve for $v$. Substituting $g = 9.8 \, \text{m/s}^2$ and rearranging, we get $v = \frac{y - \frac{1}{2}gt^2}{t}$. Substituting $y = 78.4$, $g = 9.8$, and $t = 2$, we find $v = \frac{78.4 - \frac{1}{2} \times 9.8 \times 2^2}{2} = \frac{78.4 - 19.6}{2} = \frac{58.8}{2} = 29 \, \text{m/s}$.

Ball Is Dropped at Time t=0.0 s\mathrm{t}=0.0 \mathrm{~s}t=0.0 s At Dropped t=2.0 s\mathrm{t}=2.0 \mathrm{~s}t=2.0 s

Ball Is Dropped at Time $\mathrm{t}=0.0 \mathrm{~s}$ At Dropped $\mathrm{t}=2.0 \mathrm{~s}$