Jeron has a tendency to drop his bowling ball upon release when bowling. Instead of having the ball on the floor at the completion of his swing, Jeron lets go of the ball 0.35m above the floor. If he throws the ball with a horizontal velocity of 5.3m/s, what distance does it travel before you hear a "thud"?
To find the distance the ball travels before you hear a "thud", we need to determine the time it takes for the ball to hit the floor.
We have the initial vertical position of the ball, which is 0.35m above the floor, and the vertical acceleration due to gravity, which is 9.8m/s². We can use these values to find the time it takes for the ball to reach the floor.
We can use the kinematic equation for vertical motion:
d = vi * t + (1/2) * a * t^2
where:
- d is the displacement,
- vi is the initial velocity,
- a is the acceleration, and
- t is the time.
Since the initial vertical velocity of the ball is 0 (as it is just beginning to fall), the equation simplifies to:
d = (1/2) * a * t^2
Plugging in the values, we have:
0.35 = (1/2) * 9.8 * t^2
0.35 = 4.9 * t^2
To find the time, we solve for t:
t^2 = 0.35 / 4.9
t^2 = 0.0714
Taking the square root of both sides, we get:
t ≈ 0.267 seconds
Now that we know the time it takes for the ball to hit the floor, we can find the horizontal distance traveled by the ball.
The horizontal velocity of the ball remains constant throughout its flight, so we can use the equation:
d = v * t
where:
- d is the distance,
- v is the velocity, and
- t is the time.
Plugging in the values, we have:
d = 5.3 * 0.267
d ≈ 1.413 meters
Therefore, the ball travels approximately 1.413 meters before you hear a "thud".