A football is thrown horizontally with an initial velocity of (16.5 m/s). Ignoring air resistance, the average acceleration of the football over any period of time is (-9.81 m/s2).
(a) Find the velocity vector of the ball 1.62 s after it is thrown.
(b) Find the magnitude and direction of the velocity at this time.
I started by using avg.acceleration=(Vf-Vi)/t. I calculated .375m/s. Then I realized I need ^x and ^y. How do I find that?
No, the acceleration here is downward, perpendicular to the velocity.
The veloicty keeps its initial horizontal velocity, but gains a vertical velocity.
Vvertical= acceleration*time
So add the Vvertical with Vhorizontal, to get the vector velocity.
Well, I must say, that football really knows how to defy gravity! Let's get down to business though.
(a) Since the football is thrown horizontally, the initial vertical velocity is zero, and the horizontal velocity remains constant. So, after 1.62 seconds, the vertical velocity component of the football would be:
Vvertical = acceleration * time
Vvertical = -9.81 m/s² * 1.62 s
Now, since we want the velocity vector of the ball, we need to combine the horizontal and vertical components. The horizontal component remains the same as the initial velocity (since there's no horizontal acceleration), so the velocity vector is:
Velocity = (Vhorizontal, Vvertical)
Velocity = (16.5 m/s, -9.81 m/s² * 1.62 s)
(b) To find the magnitude and direction of the velocity at this time, we can use the Pythagorean theorem and some good old trigonometry. The magnitude of the velocity can be calculated as:
Magnitude = √(Vhorizontal² + Vvertical²)
The direction can be found using:
Direction = tan^(-1)(Vvertical / Vhorizontal)
So, grab your calculator and plug in the values to find the magnitude and direction!
To find the velocity vector of the ball 1.62 s after it is thrown, you can break down the problem into horizontal and vertical components.
(a) Horizontal Component:
Since the football is thrown horizontally, its horizontal velocity remains constant. So the horizontal component of the velocity vector will be equal to its initial horizontal velocity, which is 16.5 m/s.
(b) Vertical Component:
The vertical component of the velocity is affected by the acceleration due to gravity. In this case, the acceleration is -9.81 m/s^2, which is directed downward. Therefore, the vertical component of the velocity will increase at a rate of 9.81 m/s^2.
To calculate the vertical component of the velocity, use the equation:
Vvertical = acceleration * time
Given that the time is 1.62 seconds and the acceleration is -9.81 m/s^2, you can calculate the vertical component of the velocity:
Vvertical = -9.81 m/s^2 * 1.62 s = -15.92 m/s (negative sign indicates downward direction)
To find the total velocity vector, add the horizontal and vertical components of the velocity:
Velocity vector = (16.5 m/s)î + (-15.92 m/s)ĵ
Now you have the velocity vector in terms of its x-component (horizontal) and y-component (vertical). The letter "î" denotes the unit vector in the x-direction, and "ĵ" denotes the unit vector in the y-direction.
(b) Magnitude and Direction of Velocity:
To find the magnitude of the velocity, use the Pythagorean theorem:
Magnitude of velocity = √(Vx^2 + Vy^2)
Substituting the values, you get:
Magnitude of velocity = √((16.5 m/s)^2 + (-15.92 m/s)^2) ≈ 22.62 m/s
To find the direction of the velocity, you can use trigonometry. The direction can be expressed as an angle with respect to a horizontal line. You can calculate this angle using the inverse tangent function:
Direction of velocity = arctan(Vy/Vx)
Substituting the values, you get:
Direction = arctan((-15.92 m/s) / (16.5 m/s)) ≈ -46.96° (negative sign indicates downward direction)
So, the magnitude of the velocity at 1.62 s is approximately 22.62 m/s, and the direction is approximately -46.96° downward from the horizontal.