Answer:
D. 10.3 m/s
Explanation:
A:
u = 0 m/s
a = 9.8 m/s²
t = 10 s
Using
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ s = \frac{1}{2} (9.8) {(10)}^{2} \\ s = 490m[/tex]
B:
Since B is thrown one second after dropping A, B must take 9 seconds to reach the ground simultaneously with A.
s = 490m
a = 9.8 m/s²
t = 9s
Using
[tex]s = ut + \frac{1}{2} a {t}^{2} \\ u = 10.3m {s}^{ - 1} [/tex]