Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is 26.0 m/s and that the second arrow is fired 1.24 s after the first. Determine the initial speed of the second arrow.

Same procedure as 1st prob. below.

To solve this problem, we can use the kinematic equations of motion. In particular, we can use the equation that relates the final velocity, initial velocity, acceleration, and time:

v_f = v_i + at

where:
- v_f is the final velocity
- v_i is the initial velocity
- a is the acceleration
- t is the time

Let's assume that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts downward).

For the first arrow, we know that:
- v_i1 = 26.0 m/s
- t_1 is the time it takes for the first arrow to reach its maximum height

Since both arrows reach their maximum heights at the same instant, we can assume that the time it takes for the second arrow to reach its maximum height is t_1 - 1.24 s (since it is fired 1.24 s later).

Using the equation for the first arrow:
0 = 26.0 - 9.8t_1
t_1 = 26.0/9.8

Using the equation for the second arrow:
0 = v_i2 - 9.8(t_1 - 1.24)

Rearranging the equation:
v_i2 = 9.8(t_1 - 1.24)

Substituting the value of t_1:
v_i2 = 9.8(26.0/9.8 - 1.24)

Simplifying the equation:
v_i2 = 26.0 - 9.8(1.24)

Calculating:
v_i2 = 26.0 - 12.152

Final answer:
The initial speed of the second arrow is approximately 13.848 m/s.