You throw a water balloon hoping to hit your sibling. You throw it at an angle of 21.0° at 5.50 m/s. It ends up leaving your hand at a height of 1.70 m above the ground. It completely misses your sibling and breaks on the ground. How far did the water balloon travel?
in the vertical:
hf=hi+vi'*t-4.0t^2
0=1.7+5.5*sin21*t-4.9t^2
it is a quadratic equation solve for time t.
Horizontal distance:
distance=5.5cos21*t
Lucky kid
Well, it seems like your water balloon had a case of stage fright and decided to go on its own adventure! Let's calculate how far it traveled.
To determine the horizontal distance traveled, we first need to find the time of flight. We can use the equation: t = 2 * V₀ * sin(θ) / g, where V₀ is the initial velocity (5.50 m/s), θ is the launch angle (21.0°), and g is the acceleration due to gravity (approximately 9.8 m/s²).
Plugging in the values, we get:
t = 2 * 5.50 * sin(21.0°) / 9.8
After calculating, the time of flight turns out to be approximately 0.684 seconds.
Next, let's find the horizontal distance traveled. We can use the equation: d = V₀ * cos(θ) * t.
Plugging in the values, we get:
d = 5.50 * cos(21.0°) * 0.684
After calculating, the horizontal distance traveled by the water balloon is approximately 3.34 meters.
So, your water balloon managed to travel 3.34 meters, proving that sometimes it's better at exploring than hitting siblings.
To calculate the distance traveled by the water balloon, we first need to find its horizontal and vertical components of velocity.
Given:
Initial velocity (v₀) = 5.50 m/s
Launch angle (θ) = 21.0°
The horizontal component of velocity (vₓ) remains constant throughout the motion because there is no horizontal acceleration. The vertical component of velocity (vᵧ) changes due to the acceleration due to gravity.
To find the horizontal component of velocity (vₓ):
vₓ = v₀ * cos(θ)
vₓ = 5.50 m/s * cos(21.0°)
vₓ ≈ 5.084 m/s
To find the vertical component of velocity (vᵧ):
vᵧ = v₀ * sin(θ)
vᵧ = 5.50 m/s * sin(21.0°)
vᵧ ≈ 1.914 m/s
Now, we can use the vertical component of velocity (vᵧ) to calculate the time of flight of the water balloon.
Using the equation:
Vertical displacement (Δy) = vᵧ * t - (1/2) * g * t²
Where:
Vertical displacement (Δy) = -1.70 m (as it is falling down)
Acceleration due to gravity (g) = 9.8 m/s²
-1.70 = 1.914 * t - (1/2) * 9.8 * t²
-1.70 ≈ 1.914t - 4.9t²
Rearranging the equation:
4.9t² - 1.914t - 1.70 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Where:
a = 4.9
b = -1.914
c = -1.70
t₁ = [(-(-1.914) + √((-1.914)² - 4(4.9)(-1.70))) / 2(4.9)]
t₂ = [(-(-1.914) - √((-1.914)² - 4(4.9)(-1.70))) / 2(4.9)]
t₁ ≈ 0.679 s
t₂ ≈ 0.428 s
Since the time of flight cannot be negative, we take t = 0.679 s as the valid answer.
Now, we can calculate the horizontal distance (x) traveled by the water balloon using the horizontal component of velocity (vₓ) and the time of flight (t).
x = vₓ * t
x = 5.084 m/s * 0.679 s
x ≈ 3.454 m
Therefore, the water balloon traveled approximately 3.454 meters.
To determine how far the water balloon traveled, we can break down the motion into horizontal and vertical components.
First, let's find the time it takes for the balloon to hit the ground. We can use the vertical motion equation:
y = y0 + v0y * t - (1/2) * g * t^2
where:
y = final height (0 m, as it hits the ground)
y0 = initial height (1.70 m)
v0y = initial vertical velocity (5.50 m/s * sin(21.0°))
g = acceleration due to gravity (9.8 m/s^2)
t = time
Since the balloon hits the ground, we can set y = 0 and solve for t:
0 = 1.70 + (5.50 * sin(21.0°)) * t - (1/2) * 9.8 * t^2
Simplifying the equation:
4.9t^2 - (5.50 * sin(21.0°))t - 1.70 = 0
Using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = 4.9, b = -(5.50 * sin(21.0°)), c = -1.70
Solving for t, we find two roots: t = 0.34 s (ignoring the negative value)
Now, let's find the horizontal distance the balloon traveled:
x = v0x * t
where:
x = horizontal distance
v0x = initial horizontal velocity (5.50 m/s * cos(21.0°))
t = time (0.34 s)
Plugging in the values:
x = (5.50 * cos(21.0°)) * 0.34
Calculating the value, we find:
x ≈ 1.71 meters
Therefore, the water balloon traveled approximately 1.71 meters before hitting the ground.