Question

An arrow was shot straight up with an initial seed of 50.0m/s. what was the maximum height that the arrow flew?

Answer 1

use the equations of motion, specifically

\[v^2 = u^2 + 2 \ a \ x\]

\[\rightarrow x = \frac{v^2 - u^2}{2a}\]

u = 50, v = 0, a = -10

Answer 2

how is a = -10? shouldn't it be -9.8 m/s^2?

Answer 3

Yes, but in order to simplify the calculation you can say that a = -9.8 \(\sf m/s^2\) \(\approx\) -10 \(\sf m/s^2\)

Answer 4

yep the above equation will lead you to your answer