Final answer:
The initial velocity of the arrow shot straight up was 42.336 m/s, and it reached a maximum height of 91.53216 m using the kinematic equations of projectile motion with an acceleration due to gravity of -9.8 m/s^2.
Step-by-step explanation:
To solve for the initial velocity and maximum height of an arrow shot straight up, we can use kinematic equations for projectile motion, assuming acceleration due to gravity (g) is -9.8 m/s2 and air resistance is negligible.
Since the arrow travels up and then down in 8.64 seconds, the time to reach the maximum height is half of that, which is 4.32 seconds.
The initial velocity (vi) can be found using the equation:
vf = vi + at,
where vf is the final velocity (0 m/s at the maximum height), a is the acceleration due to gravity, and t is the time.
Thus, vi = -(-9.8 m/s2)(4.32 s)
= 42.336 m/s.
The maximum height (h) can be calculated using the equation:
h = vit + ½at2,
which gives h = (42.336 m/s)(4.32 s) + ½(-9.8 m/s2)(4.32 s)2
= 91.53216 m.