Problem: A hunter shoots an arrow at a deer directly away from him. When the arrow leaves the bow, the deer is at a distance of 38 m. When the arrow strikes, the deer is at a distance of 51.00 m. The speed of the arrow is 65.00 m/s. What must have been the speed of the deer? How long did the arrow take to travel to the deer?

My teacher said to calculate the time of flight to get how long the arrow traveled for to get to the deer, but I don't angle of the arrow (unless its 180 degrees).

Formula:t = (65 m/s * sin (180))/(9.81 m/s^2)

If you don't shoot the arrow at an angle,it will NOT hit the deer, but hit the ground.

The horizontal speed of the arrow is 65cosTheta
The vertical speed is 65sinTheta.

Horizontal distance= horizvelocity*time
or time= 51/65cosTheta

Now put that time here:
Hfinal=Hinitial+ Vi*t -1/2 g t^2
but Hfinal=Hinitial,so
0=65sintheta *t - 4.9t^2
Now put the expression for t you got above, and solve for theta (it is a little messy, and you get two solutions).
Then go back and solve for time.

To calculate the speed of the deer, we can use the equation of motion:

Speed of arrow = Speed of deer + Speed of arrow relative to deer

Let's assume the speed of the deer is v_deer.

Given:
Speed of arrow = 65.00 m/s
Distance when arrow leaves bow (initial distance) = 38 m
Distance when arrow strikes (final distance) = 51.00 m

We know that the arrow moves directly away from the hunter, so the relative speed of the arrow with respect to the deer is the difference in distances divided by the time taken:

Relative speed = (final distance - initial distance) / time

Substituting the given values, we get:

Relative speed = (51.00 m - 38.00 m) / time

Now, to find the time taken by the arrow to reach the deer, we can calculate the time of flight. The time of flight can be calculated by dividing the initial distance by the speed of the arrow. Since the arrow is shot directly away from the hunter, its initial speed will not change during its flight. Therefore, the time taken by the arrow to reach the deer is:

Time taken by arrow = initial distance / speed of arrow
= 38.00 m / 65.00 m/s

Now, we can substitute this value of time back into the relative speed equation:

Relative speed = (51.00 m - 38.00 m) / (38.00 m / 65.00 m/s)

Finally, to find the speed of the deer, we rearrange the equation:

Speed of deer = Speed of arrow - Relative speed

Substituting the given values, we get:

Speed of deer = 65.00 m/s - Relative speed

Calculating the relative speed and substituting it back into the equation, you can find the speed of the deer.

t = 38.00 m / 65.00 m/s (you can calculate this value)
Relative speed = (51.00 m - 38.00 m) / t (substituting the value of t)
Speed of deer = 65.00 m/s - Relative speed (calculating speed of the deer)

The formula you mentioned, t = (65 m/s * sin(180))/(9.81 m/s^2), seems to be incorrect for calculating the time of flight. It appears to be attempting to calculate the time using the horizontal component of the arrow's initial velocity, which is not necessary in this case since the problem specifies that the arrow is shot directly away from the hunter.

To solve this problem, let's use the equation of motion for the arrow:

s = ut + (1/2) * at^2

where:
s is the displacement of the arrow (51 m),
u is the initial velocity of the arrow (65 m/s),
t is the time taken,
and a is the acceleration (which is -9.81 m/s^2 since the arrow is moving opposite to gravity).

Simplifying the equation, we get:

51 m = 65 m/s * t + (1/2) * (-9.81 m/s^2) * t^2

Rearranging and simplifying further, we have:

0.5 * (-9.81 m/s^2) * t^2 + 65 m/s * t - 51 m = 0

Now, we can solve this quadratic equation for t using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 0.5 * (-9.81 m/s^2), b = 65 m/s, and c = -51 m. Plugging in these values, we can calculate the time t.

Let's plug the values into the quadratic formula and solve for t:

t = (-65 m/s ± √((65 m/s)^2 - 4 * (0.5 * (-9.81 m/s^2)) * (-51 m))) / (2 * 0.5 * (-9.81 m/s^2))

Calculating the expression under the square root:

√((65 m/s)^2 - 4 * (0.5 * (-9.81 m/s^2)) * (-51 m) )
= √(4225 m^2/s^2 - 4 * (0.5 * (-9.81 m/s^2)) * (-51 m) )
= √(4225 m^2/s^2 - 20.8 m/s^2 * (-51 m))
= √(4225 m^2/s^2 + 1060.8 m^2/s^2)
= √(5285.8 m^2/s^2)
≈ 72.68 m/s

Now, substituting this value into the quadratic formula:

t = (-65 m/s ± 72.68 m/s) / (-9.81 m/s^2)

Calculating the two possible values of t:

t1 = (-65 m/s + 72.68 m/s) / (-9.81 m/s^2) ≈ -0.76 seconds
t2 = (-65 m/s - 72.68 m/s) / (-9.81 m/s^2) ≈ 14.16 seconds

Since time cannot be negative, we discard t1 and keep t2 as the valid solution.

Therefore, the arrow took approximately 14.16 seconds to travel to the deer.

To calculate the speed of the deer, we can use the formula:

speed of deer = distance traveled by deer / time taken by arrow

The distance traveled by the deer is the difference between the initial distance and the final distance: 51 m - 38 m = 13 m.

Substituting the values, we have:

speed of deer = 13 m / 14.16 s ≈ 0.92 m/s

Therefore, the speed of the deer is approximately 0.92 m/s.