vIRGINIA THREW A fRISBEE FOR A DISTANCE OF 138.56 M TO CAPTURE THE WOMEN'S RECORD. IF THE FRISBEE WAS THROWN HORIZONTALLY WITH A SPEED OF 13.0M/S HOW LON DID THE FRISBEE REMAIN ALOFT?
sine its horizontal speed remained constant, it took 138.56/13.0 seconds to hit the ground.
To calculate how long the frisbee remained aloft, we can use the equation of motion for projectile motion. In this case, since the frisbee was thrown horizontally, its initial vertical velocity is 0 m/s. We can use the equation:
distance = (initial velocity x time) + (1/2 x acceleration x time^2)
Since the initial vertical velocity is 0 m/s, the equation simplifies to:
distance = (1/2 x acceleration x time^2)
In this case, the vertical distance traveled by the frisbee is equal to the height of its trajectory, which is 0 m since it was thrown horizontally.
So the equation becomes:
0 = (1/2 x acceleration x time^2)
Since acceleration due to gravity is approximately 9.8 m/s^2, we can rewrite the equation as:
0 = (1/2 x 9.8 x time^2)
Simplifying further:
0 = 4.9 x time^2
Now, we can solve for time by rearranging the equation:
time^2 = 0 / 4.9
time^2 = 0
Therefore, the frisbee remained aloft for 0 seconds since it was thrown horizontally.