4.if cotθ� = -√15/7 , use the fundamental identities to find cosθ� > 0
9.A projectile is fired with an initial velocity of 88 ft/sec at an angle of 35° to the horizontal. Neglecting air resistance, what is the time of impact? (Round to the nearest tenth of a second.)
Remember: x = (v↧0cosθ�)t and y = h↧0+ (v↧0sinθ�)t – 16t^2
4.
√(15+49) = 8, so cosθ = 7/8
9.
we want when y=0
y = 88sin35° t - 16t^2
y=0 when t = 88sin35°/32 = 1.57
so, t = 1.6 sec
The answer for number 4 is wrong, Cotangent means adj/opp so it would be -15/8
To find the value of cosθ, given cotθ = -√15/7, we can use the fundamental identity relating cotangent and tangent:
cot²θ + 1 = csc²θ
Rearranging the equation and substituting the given value for cotθ:
cot²θ = csc²θ - 1
(√15/7)² = (1/sinθ)² - 1
15/49 = 1/sin²θ - 1
15/49 + 1 = 1/sin²θ
(15 + 49)/49 = 1/sin²θ
64/49 = 1/sin²θ
Taking the reciprocals on both sides:
49/64 = sin²θ
Taking the square root on both sides:
sqrt(49/64) = sqrt(sin²θ)
7/8 = sinθ
Now, to find cosθ, we can use the Pythagorean identity:
sin²θ + cos²θ = 1
Substituting the value of sinθ we found:
(7/8)² + cos²θ = 1
49/64 + cos²θ = 1
cos²θ = 1 - 49/64
cos²θ = 15/64
Taking the square root on both sides:
sqrt(cos²θ) = sqrt(15/64)
|cosθ| = sqrt(15)/8
Since cosθ represents the cosine of an angle, it can only take values between -1 and 1. Therefore, cosθ must be positive. Thus, cosθ > 0, or simply cosθ is positive.
To find the time of impact for the projectile fired at an angle of 35° to the horizontal, we can use the equation for vertical motion:
y = h₀ + (v₀sinθ)t - 16t²
Where:
- y is the vertical displacement (in this case, it would be the height above the ground, or h₀)
- h₀ is the initial height (in this case, it is 0 since we assume the projectile is launched from the ground-level)
- v₀ is the initial velocity (given as 88 ft/sec)
- θ is the angle of launch (given as 35°)
- t is the time of flight (which we are trying to find)
Since we are looking for the time when the projectile impacts the ground, the vertical displacement (y) would be 0.
0 = 0 + (88sin35°)t - 16t²
Simplifying the equation:
0 = (88 × 0.574)t - 16t²
0 = 50.512t - 16t²
Setting the equation equal to zero:
16t² - 50.512t = 0
Factoring out a t:
t(16t - 50.512) = 0
Setting each factor equal to zero:
t = 0 (which is not valid in this case)
16t - 50.512 = 0
Solving for t:
16t = 50.512
t = 50.512/16
t ≈ 3.157
Therefore, the time of impact is approximately 3.2 seconds (rounded to the nearest tenth).