1.If an arrow is shot upward on the moon with a velocity of 50 m/s its height (in meters) after t seconds is given by H = 50t - 0.66t2. With what velocity will the arrow hit the moon?

a. -49.5
b. -47
c. -50.1
d. -50
e. -51
2.If an arrow is shot upward on the moon with a velocity of 53 m/s its height (in meters) after t seconds is given by H = 53t - 0.62t2. When will the arrow hit the moon? Round the result to the nearest thousandth if necessary.
a. 85.484
b. 85.473
c. 85.489
d. 85.374
e. 85.485

clearly, due to conservation of energy, etc., the arrow will strike with the same velocity as it left, but in the opposite direction.

just solve 53t - 0.62 t^2 = 0
t(53-0.62t) = 0
t = 53/0.62 = 85.4839

To find the velocity at which the arrow hits the moon in the first problem, we need to find the value of t when the height (H) is zero.

1. Set H = 0:
0 = 50t - 0.66t^2

2. Rearrange the equation:
0.66t^2 - 50t = 0

3. Factor out t:
t(0.66t - 50) = 0

4. Set each factor equal to zero to find the values of t:
t = 0 (corresponding to when the arrow is shot)
0.66t - 50 = 0
0.66t = 50
t ≈ 75.76

So, the arrow will hit the moon at approximately t = 75.76 seconds.

To find the velocity at which the arrow hits the moon in the second problem, we again set H = 0:

1. Set H = 0:
0 = 53t - 0.62t^2

2. Rearrange the equation:
0.62t^2 - 53t = 0

3. Factor out t:
t(0.62t - 53) = 0

4. Set each factor equal to zero to find the values of t:
t = 0 (corresponding to when the arrow is shot)
0.62t - 53 = 0
0.62t = 53
t ≈ 85.484

So, the arrow will hit the moon at approximately t = 85.484 seconds.

Therefore, the answers to the questions are:
1. The arrow will hit the moon with a velocity of approximately -50 m/s.
(The answer is option d. -50)
2. The arrow will hit the moon after approximately 85.484 seconds.
(The answer is option a. 85.484)

1. To find the velocity at which the arrow hits the moon, we need to find the derivative of the height equation with respect to time.

H = 50t - 0.66t^2

Differentiating both sides with respect to t:

dH/dt = 50 - 1.32t

Setting dH/dt to 0 (since the arrow hits the moon at its peak height):

0 = 50 - 1.32t

Solving for t:

1.32t = 50
t = 50 / 1.32
t ≈ 37.878

Now, substituting this value of t back into the velocity equation to find the velocity at this time:

v = 50 - 1.32t
v = 50 - 1.32(37.878)
v ≈ -49.915

Rounding to the nearest tenth, the velocity at which the arrow hits the moon is approximately -49.9 m/s.

Therefore, the closest answer choice is (c) -50.1 m/s.

Answer: c. -50.1

2. Similar to the previous question, we need to find the time when the height of the arrow is zero (when it hits the moon).

H = 53t - 0.62t^2

Setting H to zero and solving for t:

0 = 53t - 0.62t^2
0 = t(53 - 0.62t)

Either t = 0 (at the beginning) or 53 - 0.62t = 0.

Solving 53 - 0.62t = 0 for t:

0.62t = 53
t = 53 / 0.62
t ≈ 85.484

Rounding to the nearest thousandth, the arrow will hit the moon approximately at t = 85.484 seconds.

Therefore, the answer is (a) 85.484.

Answer: a. 85.484