a projectile with a horizontal component of 13m/s and a vertical component of 18 m/s would have an overall velocity of m/s??? at and angle of ?? above the horizontal.
Can you please help me fill in the two blanks thank you
x = 13 m/s
y = 18 m/s
r = √( 13² + 18² ) = 22.2 m/s
angle = tan⁻¹ (18/13) = 54.2°
Sure, I'd be happy to help! To find the overall velocity and angle of the projectile, we can use the Pythagorean theorem and trigonometric functions.
Given that the horizontal component of velocity is 13 m/s and the vertical component is 18 m/s, we can use the Pythagorean theorem (c^2 = a^2 + b^2) to find the overall velocity (c):
c^2 = (13 m/s)^2 + (18 m/s)^2
c^2 = 169 m^2/s^2 + 324 m^2/s^2
c^2 = 493 m^2/s^2
Taking the square root of both sides, we find that c ≈ 22.2036 m/s. Therefore, the overall velocity is approximately 22.2036 m/s.
To find the angle above the horizontal (θ), we can use the inverse tangent (arctan) function:
θ = arctan(vertical component / horizontal component)
θ = arctan(18 m/s / 13 m/s)
θ ≈ arctan(1.3846)
Using a calculator, we can find that θ ≈ 54.06 degrees (rounded to two decimal places). Therefore, the angle above the horizontal is approximately 54.06 degrees.
So, the overall velocity is approximately 22.2036 m/s, at an angle of approximately 54.06 degrees above the horizontal.
To find the overall velocity of the projectile, we can use the Pythagorean theorem. The horizontal and vertical components form a right triangle, where the overall velocity is the hypotenuse.
Using the given values:
Horizontal component (Vx) = 13 m/s
Vertical component (Vy) = 18 m/s
We can calculate the overall velocity (V) as follows:
V = sqrt(Vx^2 + Vy^2)
Substituting the given values:
V = sqrt((13 m/s)^2 + (18 m/s)^2)
V = sqrt(169 m^2/s^2 + 324 m^2/s^2)
V = sqrt(493 m^2/s^2)
V ≈ 22.2 m/s
So, the overall velocity of the projectile is approximately 22.2 m/s.
To find the angle above the horizontal, we can use trigonometry. The angle (θ) can be determined using the inverse tangent function:
θ = arctan(Vy / Vx)
Substituting the given values:
θ = arctan(18 m/s / 13 m/s)
θ = arctan(1.38)
θ ≈ 54.6 degrees
Therefore, the overall velocity of the projectile is approximately 22.2 m/s, and the angle above the horizontal is approximately 54.6 degrees.
To find the overall velocity of the projectile, we can use the Pythagorean theorem because the horizontal and vertical components of velocity form a right triangle. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the horizontal component of velocity is 13 m/s and the vertical component is 18 m/s. Therefore, we have:
Overall velocity^2 = horizontal velocity^2 + vertical velocity^2
Plugging in the given values, we have:
Overall velocity^2 = 13 m/s^2 + 18 m/s^2
Overall velocity^2 = 169 m^2/s^2 + 324 m^2/s^2
Overall velocity^2 = 493 m^2/s^2
Taking the square root of both sides, we get:
Overall velocity = √493 m/s
Simplifying this, we find that the overall velocity of the projectile is approximately 22.2 m/s.
Now, let's calculate the angle above the horizontal. We can use the trigonometric ratios to determine the angle. By dividing the vertical component of velocity by the horizontal component of velocity, we get the tangent of the angle:
tan(angle) = vertical velocity / horizontal velocity
Substituting the given values, we have:
tan(angle) = 18 m/s / 13 m/s
tan(angle) = 1.3846
To find the angle, we can use the inverse tangent (arctan or tan^(-1)). Taking the inverse tangent of both sides, we get:
angle = arctan(1.3846)
Calculating this using a calculator, we find that the angle is approximately 53.1 degrees.
Therefore, the overall velocity of the projectile is approximately 22.2 m/s at an angle of approximately 53.1 degrees above the horizontal.