You throw a water balloon hoping to hit your sibling. You throw it at an angle of 28.0° at 9.60 m/s. It ends up leaving your hand at a height of 2.00 m above the ground. It completely misses your sibling and breaks on the ground. How far did the water balloon travel?
The height y = 2 + (9.6 sin28°)t - 4.9t^2
find when y=0, then use that t to find the distance
x = 9.6 cos28° * t
To determine how far the water balloon traveled, we can use the principles of projectile motion. The horizontal and vertical motions can be treated as separate one-dimensional motions.
First, we need to find the time it takes for the water balloon to hit the ground. We can use the vertical motion equation:
h = (v₀y * t) + (1/2 * g * t²),
where:
h = initial height = 2.00 m
v₀y = initial vertical velocity (y-component) = 9.60 m/s * sin(28.0°)
g = acceleration due to gravity = 9.8 m/s²
t = time taken to reach the ground
Rearranging the equation and substituting the known values, we get:
0 = (4.3596 * t) + (0.5 * 9.8 * t²)
Simplifying this equation, we obtain a quadratic equation:
4.9 * t² + 4.3596 * t - 2 = 0
Solving this equation, we find two possible values for t: t = -0.6187 s and t = 0.4463 s. Since time cannot be negative, we discard the negative value. Therefore, it takes approximately 0.4463 seconds for the water balloon to hit the ground.
Next, we can find the horizontal distance traveled by the water balloon using the horizontal motion equation:
d = v₀x * t,
where:
d = horizontal distance traveled
v₀x = initial horizontal velocity (x-component) = 9.60 m/s * cos(28.0°)
t = time taken to reach the ground = 0.4463 s
Substituting the known values, we can calculate the horizontal distance:
d = (9.60 * cos(28.0°)) * 0.4463
Evaluating this expression, we find that the water balloon traveled approximately 3.952 meters before hitting the ground and breaking.