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You are going to launch a water balloon at 3.5 m/s at an angle of 35o . how high does the water balloon in go (δy)?

User Xevincent
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Final answer:

To find the height the water balloon reaches, we can use the kinematic equations for projectile motion. By using the formula for maximum height and substituting the given values, we can calculate that the water balloon reaches a height of approximately 0.41 meters.

Step-by-step explanation:

To calculate the maximum height reached by the water balloon, we can use the kinematic equations for projectile motion. The equation we need to use is:

delta(y) = (v0y2sin^2(theta)) / 2g

In this equation, delta(y) represents the maximum height, v0y is the initial vertical velocity, theta is the launch angle, and g is the acceleration due to gravity.

Given that the initial velocity (v0) is 3.5 m/s and the launch angle (theta) is 35 degrees, we can find the initial vertical velocity using the equation:

v0y = v0sin(theta)

Substituting the values into the equation, we have:

v0y = 3.5 m/s * sin(35 degrees)

Solving for v0y, we find v0y = 2.0 m/s.

Now we can substitute the values into the equation for delta(y), giving:

delta(y) = (2.0 m/s)^2 * sin^2(35 degrees) / (2 * 9.8 m/s^2)

Calculating this expression, we find that the water balloon reaches a maximum height of approximately 0.41 meters.

User KrisG
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