An arrow is shot upward on mars from ground level with a velocity of 60 meters per second. its height (in meters) after t seconds is given by 2 s(t) = 60t -1.9t^2 . (a) what is the maximum height attained by the arrow? (b) with what velocity will the arrow hit mars?

max height = 473.68 m

velocity hit mars = 60 m/s

Explanation:

Maximum height can be found by finding the 1st derivative of s(t) and equate it to zero.

s'(t) = ds/dt = 60 - 3.8t

s't() = 0

60 - 3.8t = 0

3.8t = 60

t = 60/3.8 = 15.79

subs t = 15.79 to s(t)

s(15.79) = 60(15.79) - 1.9(15.79)^2 = 473.68 m

b) The arrow will hit mars after it went up to the maximum height and travelled back downward due to gravity

Assuming the gravity constant, the velocity when it hit the ground should be the same as it leaves the ground. To confirm that, we tested with the equation of motion.

Since there is no gravity given, let a downward as g

v^2 = u^2 + 2as

The arrow shot upward will comes back downward. Since gravity is always constant, the time it took back to reach the ground should be the same as it goes up to max height

so, t = 2*15.79 = 31.58 s

s = 0 since we are looking at the moment it touches back the ground

v^2 = u^2 +2as

v^2 = 60^2 + 2g(0)

v^2 = 60^2

v = 60 m/s

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