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 the that initial speed of the first arrow is 34 m/s and that the second arrow is fired 0.204081632653061 seconds after the first. Determine the initial speed of the second arrow.

The initial speed of the second arrow is 33.8 m/s.

Explanation:

initial speed of first arrow, u = 34 m/s

Let the initial height of the second arrow is h.

Let they both reaches at maximum height H.

Let the time taken by the first arrow is t and the second arrow is t - 0.0204

Let the initial speed of the second arrow is u'.

Use first equation of motion for the first arrow.

v = u - gt

0 = u - gt

34 = gt (1)

For the second arrow

v =u' - g (t - 0.0204)

0 = u' - gt + 9.8 x 0.0204

u' = 34 - 0.1999 = 33.8 m/s

what are the options.

explanation:

a) electrical > light, sound, and thermal< < < <

) the mechanical energy of the firefighter moving.< < < < a) electrical > light, sound, and thermal< < < <

c) both a and b.