If an arrow is shot upward on the moon with a velocity of 58 m/s, its height (in meters) after t seconds is given by H = 58t − 0.83t2. (a) Find the velocity of the arrow after two seconds. m/s (b) Find the velocity of the arrow when t = a. m/s (c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s (d) With what velocity will the arrow hit the surface? m/s

a) v = 54.7m/s

b) v = (58 - 1.66a) m/s

c) t = 69.9 s

d) v = -58.0 m/s

Explanation:

Given;

The height equation of the arrow;

H = 58t - 0.83t^2

(a) Find the velocity of the arrow after two seconds. m/s;

The velocity of the arrow v can be given as dH/dt, the change in height per unit time.

v = dH/dt = 58 - 2(0.83t) 1

At t = 2 seconds

v = dH/dt = 58 - 2(0.83×2)

v = 54.7m/s

(b) Find the velocity of the arrow when t = a. m/s

Substituting t = a into equation 1

v = 58 - 2(0.83×a)

v = (58 - 1.66a) m/s

(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s

the time when H = 0

Substituting H = 0, we have;

H = 58t - 0.83t^2 = 0

0.83t^2 = 58t

0.83t = 58

t = 58/0.83

t = 69.9 s

(d) With what velocity will the arrow hit the surface? m/s

from equation 1;

v = dH/dt = 58 - 2(0.83t)

Substituting t = 69.9s

v = 58 - 2(0.83×69.9)

v = -58.0 m/s

the part of the wave he is travelling on is the crest

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this