Question

a person drops a ball (a) out a window and also throws a ball (b) down at the same time with an initial velocity of 10 m/s.
a) what is the velocity of a after 2 seconds?
b) how far has a fallen in 2 seconds?
c) if ball b has a velocity of 40 m/s, how long ago was it thrown downward?

Explanation:

Step1: Find velocity of A

Use the formula \(v = v_0+at\). For ball A, \(v_0 = 0\ m/s\) (dropped), \(a = g= 9.8\ m/s^2\), \(t = 2\ s\).
\(v=0 + 9.8\times2\)

Step2: Find distance fallen by A

Use the formula \(s=v_0t+\frac{1}{2}at^2\). For ball A, \(v_0 = 0\ m/s\), \(a = g = 9.8\ m/s^2\), \(t = 2\ s\).
\(s=0\times2+\frac{1}{2}\times9.8\times2^{2}\)

Step3: Find time for ball B

Use the formula \(v = v_0+at\). For ball B, \(v = 40\ m/s\), \(v_0 = 10\ m/s\), \(a = g=9.8\ m/s^2\).
\(40=10 + 9.8t\), then solve for \(t\): \(9.8t=40 - 10\), \(t=\frac{30}{9.8}\)

Answer:

a) \(19.6\ m/s\)
b) \(19.6\ m\)
c) \(\approx3.06\ s\)