1) A bullet is found embedded in the wall of a room 2.9 m above the floor. The bullet entered the wall going upward at an angle of 39.6°. How far from the wall was the bullet fired if the gun was held 1.1 m above the floor?
2) Let sin θ = 2/7. Find the exact value of cos θ.
for question 2, I did
cos (θ) =sin(90degrees −θ)
cos (θ) = sin (90 - 2/7)
cos (θ) = 0.999987566
1. I don't see how the 1.1 m has anything to do with the problem
I simply see:
cot 39.6° = x/2.9
x = 2.9 cot 39/6° or 2.9/tan39.6° = ...
2. first of all, when you did sin(90° - 2/7)
you really found sin(90° - sin θ), which is incorrect
given sin θ = 2/7 which is y/r , let's make y = 2, and r = 7
we know x^2 + y^2 = r^2
x^2 + 4 = 49
x = ± √45 = ± 3√5
so cos θ = 3√5/7 or -3√5/7 , you didn't say where θ was.
for question 1, I got 3.5055
for question 2, I entered -3√5/7 in my calculator and I got -2.5355
is my answer correct?
I also got 3.505 for the first question,
but....
Whenever it asks for exact values, they expect either exact fractions or
exact square roots,
that is why the correct exact value is ± 3√5/7
besides all that (3√5)/7 = appr .95831.... not 2.5355
These kind of questions can be checked on a calculator
press the following sequence: (my calculator is a scientific SHARP)
make sure it is set to DEG
press: 2 ÷7 =
you should get .2857...
press: 2ndF sin =
you should get 16.6015... °
now press: cos =
and you will get .95831..
now do:
- 3 x √5 ÷ 7 =
you should get 0 , showing that cosθ is indeed 3√5/7 exactly!!!
ok thank you
1) To solve this problem, we can use the principles of projectile motion. We know that the bullet entered the wall at an angle of 39.6° with respect to the horizontal and the gun was held 1.1 m above the floor.
First, let's calculate the vertical displacement of the bullet. We know that the vertical height of the bullet in the wall is 2.9 m above the floor, and the gun was held 1.1 m above the floor. Therefore, the bullet traveled a vertical distance of (2.9 - 1.1) m = 1.8 m.
Next, we can calculate the time of flight of the bullet using the vertical displacement. The vertical motion of the bullet can be described by the equation:
y = yo + vot - (1/2)gt^2,
where y is the vertical displacement, yo is the initial vertical position (1.1 m for the gun height), vo is the initial vertical velocity, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time of flight.
Plugging in the values we have:
1.8 = 0 + vo*t - (1/2)*9.8*t^2.
Simplifying the equation gives us:
4.9t^2 - vot + 1.8 = 0.
This is a quadratic equation in terms of t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a),
where a = 4.9, b = -vo, and c = 1.8.
Since the bullet was fired upward at an angle of 39.6°, the initial vertical velocity can be calculated as vo = v * sin(θ), where v is the initial velocity of the bullet. We don't know the exact value of v yet.
Now, using the horizontal motion of the bullet, we can describe its displacement using the equation:
x = voH * t,
where x is the horizontal displacement, voH is the initial horizontal velocity, and t is the time of flight.
The initial horizontal velocity voH can be calculated as voH = v * cos(θ), where θ is the angle of entry (39.6°) and v is the initial velocity of the bullet.
Since the vertical and horizontal displacements are related by the angle of entry, we can express voH in terms of vo:
voH = vo * cos(θ) = (vo * cos(39.6°)).
Finally, we need to find the horizontal distance from the wall at which the bullet was fired. By substituting the expressions for t and voH into the equation for x, we get:
x = (vo * cos(39.6°)) * [(-b ± √(b^2 - 4ac)) / (2a)].
Now we have an equation that relates the horizontal distance x to the velocity vo of the bullet. Solving this equation will give us the answer to the problem.
2) To find the exact value of cos θ, given sin θ = 2/7, we can use the Pythagorean identity:
sin^2 θ + cos^2 θ = 1.
Since we know sin θ = 2/7, we can substitute it into the equation:
(2/7)^2 + cos^2 θ = 1.
Simplifying this equation:
4/49 + cos^2 θ = 1.
Subtracting 4/49 from both sides:
cos^2 θ = 1 - 4/49.
cos^2 θ = 45/49.
Taking the square root of both sides:
cos θ = ± √(45/49).
Since cos θ is positive in the first and fourth quadrants, the exact value becomes:
cos θ = √(45/49).
We can simplify this further by canceling out the common factors in the numerator and denominator:
cos θ = √(45)/√(49).
Since the square root of 49 is 7:
cos θ = √(45)/7.
So the exact value of cos θ is √(45)/7.