Answer:
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
