quarterback throws a pass to a receiver, who catches it at the same height as the pass us thrown, The initial velocity of the ball is 15.0 m/s, at an angle of 25.0 degrees above the horizontal. How long is the ball in the air?
Well, it sounds like we have a case of "air mail" here. To find out how long the ball is in the air, we can use some projectile motion skills. Since the question mentions that the receiver catches the ball at the same height as it's thrown, we can ignore any vertical displacement.
Now, the initial velocity of the ball is 15.0 m/s at an angle of 25.0 degrees above the horizontal. This means that we need to break down the velocity into its horizontal and vertical components.
The horizontal velocity (Vx) remains constant because there are no horizontal forces acting on the ball. So Vx = 15.0 m/s.
The vertical velocity (Vy) changes due to gravity. We can use some trigonometry to find Vy. Vy = V * sin(theta), where V is the initial velocity (15.0 m/s) and theta is the angle (25.0 degrees).
Vy = 15.0 m/s * sin(25.0 degrees) = 6.4 m/s.
Now, we can use the vertical velocity to find out how long the ball is in the air. We know that the acceleration due to gravity is acting downwards (-9.8 m/s^2). The time of flight can be calculated using the equation Vy = V0y + at.
So, 0 = 6.4 m/s + (-9.8 m/s^2) * t.
Solving for t, we get t ≈ 0.653 seconds.
So, the ball is in the air for approximately 0.653 seconds. It may not be a long time, but hey, at least the receiver has enough time to plan their touchdown celebration! Keep those passes flying!
To find out how long the ball is in the air, we can use the equation of motion for the vertical direction.
The equation is:
y = y0 + v0y * t - 0.5 * g * t^2
Where:
- y is the vertical displacement (which is zero in this case since the receiver catches the ball at the same height as it was thrown)
- y0 is the initial vertical position (also zero in this case)
- v0y is the initial vertical velocity (the vertical component of the initial velocity)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
First, we need to determine the initial vertical velocity, v0y. We can do this by finding the vertical component of the initial velocity using trigonometry:
v0y = v0 * sin(theta)
where v0 is the initial velocity of the ball (15.0 m/s) and theta is the angle above the horizontal (25.0 degrees).
v0y = 15.0 m/s * sin(25.0 degrees)
Next, we can substitute the known values into the equation of motion and solve for t:
0 = 0 + (15.0 m/s * sin(25.0 degrees)) * t - 0.5 * (-9.8 m/s^2) * t^2
Simplifying the equation:
4.9 * t^2 - 3.12 * t = 0
This equation can be factored as:
t * (4.9 * t - 3.12) = 0
Solving for t:
t = 0 (which is not relevant in this case) or t = 3.12 / 4.9
t ≈ 0.637 seconds
Therefore, the ball is in the air for approximately 0.637 seconds.