If an arrow is shot straight upward on the moon with a velocity of 73 m/s, its height (in meters) after t seconds is given by s(t) = 73 t - 0.83 t^2.
(A) What is the velocity of the arrow (include units) after 5 seconds?
(B) How long will it take for the arrow to return and hit the moon (include units)?
(C) With what velocity (include units) will the arrow hit the moon?
To answer these questions, we need to use the equation provided: s(t) = 73t - 0.83t^2, where s(t) represents the height of the arrow.
(A) What is the velocity of the arrow after 5 seconds?
To find the velocity of the arrow, we need to calculate the derivative of s(t) with respect to time (t). So, let's differentiate s(t):
s'(t) = d/dt (73t - 0.83t^2)
= 73 - 2(0.83t)
= 73 - 1.66t
Now, we substitute t = 5 into the equation to find the velocity after 5 seconds:
s'(5) = 73 - 1.66(5)
= 73 - 8.3
= 64.7
Therefore, the velocity of the arrow after 5 seconds is 64.7 m/s (meters per second).
(B) How long will it take for the arrow to return and hit the moon?
To find the time it takes for the arrow to return and hit the moon, we need to determine when the height of the arrow (s(t)) becomes zero. We set the equation s(t) = 0 and solve for t:
73t - 0.83t^2 = 0
This equation is a quadratic equation, so we can factor it or use the quadratic formula to find the roots. Factoring it, we get:
t(73 - 0.83t) = 0
So, either t = 0 or 73 - 0.83t = 0
From the first condition, t = 0, we can ignore it since it represents the starting point.
For the second condition:
73 - 0.83t = 0
0.83t = 73
t = 73 / 0.83
t ≈ 87.95
Therefore, it will take approximately 87.95 seconds for the arrow to return and hit the moon.
(C) With what velocity will the arrow hit the moon?
To determine the velocity at which the arrow hits the moon, we need to find the velocity at time t = 87.95 seconds using the derivative of s(t):
s'(t) = 73 - 1.66t
Substitute t = 87.95 into the equation:
s'(87.95) = 73 - 1.66(87.95)
= 73 - 145.787
≈ -72.787
Therefore, the velocity at which the arrow hits the moon is approximately -72.787 m/s (negative velocity indicating downward direction).