a projectile is fired at such an angel from the horizontal that the vertical componet of its velocity is 49m/s. the horizontal componet of its velocity is 61 m/s. what is the initial velocity of the projectile?
The resultant velocity of two velocities at 90° (such as x and y) apart can be obtained by
resultant=sqrt(x-component²+y-component²)
1 answer
To find the initial velocity of the projectile, you can use the concept of vector components. The horizontal and vertical components of the velocity can be represented as follows:
Horizontal component (Vx) = 61 m/s
Vertical component (Vy) = 49 m/s
The initial velocity (V) can be determined by using the Pythagorean theorem, which states that the square of the hypotenuse (V) is equal to the sum of the squares of the other two sides (Vx and Vy). In equation form, this can be written as:
V^2 = Vx^2 + Vy^2
Now, substitute the given values into the equation:
V^2 = (61 m/s)^2 + (49 m/s)^2
Calculating the equation:
V^2 = 3721 m^2/s^2 + 2401 m^2/s^2
V^2 = 6122 m^2/s^2
To solve for V, take the square root of both sides:
V = √(6122 m^2/s^2)
Using a calculator or performing the calculation, you will find:
V ≈ 78.3 m/s
Therefore, the initial velocity of the projectile is approximately 78.3 m/s.