A ball with a mass of 0.374 kg is thrown horizontally from a 0.8 m tall table at a 3.37 m/s velocity. How far will the ball fly from where it has been thrown?
Will changing the mass change the answer to this question?
h = 0.5g*t^2 = 0.8 m.
4.9t^2 = 0.8,
t = 0.404 s.
d = Xo * t = 3.37 * 0.404 =
To find out how far the ball will fly from where it has been thrown, we need to analyze the projectile motion of the ball. Since the ball is thrown horizontally, its initial vertical velocity is zero. However, we know its initial horizontal velocity is 3.37 m/s.
We can use the equation of motion for vertical displacement to find out how long the ball will be in the air. The equation is:
Δy = Vyi * t + (1/2) * g * t^2
Where:
- Δy is the vertical displacement (in this case, the height of the table, which is 0.8 m)
- Vyi is the initial vertical velocity (which is zero)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time of flight
Since the ball starts with zero vertical velocity, we can simplify the equation to:
0.8 m = (1/2) * 9.8 m/s^2 * t^2
Simplifying further, we have:
0.4 m = 4.9 m/s^2 * t^2
Dividing both sides by 4.9 m/s^2, we get:
t^2 = 0.4 m / 4.9 m/s^2
t^2 ≈ 0.0816 s^2
Taking the square root of both sides, we find:
t ≈ √(0.0816 s^2)
t ≈ 0.285 s
Now, we can calculate the horizontal displacement (range) using the equation:
Δx = Vx * t
Where:
- Δx is the horizontal displacement (range)
- Vx is the initial horizontal velocity (which is 3.37 m/s)
- t is the time of flight (which is approximately 0.285 s)
Plugging in the values, we have:
Δx = 3.37 m/s * 0.285 s
Δx ≈ 0.962 m
Therefore, the ball will fly approximately 0.962 meters from where it has been thrown.