An arrow is launched straight up from the ground with an initial velocity of 23.4 m/s. How long until it reaches its highest point

Answer 1

If you're shooting it from the ground, you can just find the total time in the air and divide by two.

We start with the equation y_{f = (-1/2)gt}² + v₀t + y₀.

Plugging in our values, we get 0 = (-1/2)(9.81)t² + 23.4t.

This can then be factored, and we can solve for each factor: 0 = t [ (-1/2)(9.81)t + 23.4 ]

The 't' outside just gives us zero, which is the initial time (i.e. the beginning of the problem), so we can ignore it.

Solving for the other factor, we get:

(-1/2)(9.81)t + 23.4 = 0

==> -4.905t + 23.4 = 0

==> -4.905t = -23.4

==> t = 4.77 seconds.

Half of this is 2.39 seconds. Tada!