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laiz [17]
3 years ago
6

In terms of the variables in the problem, determine the time, t, after the launch it takes the balloon to reach the target. Your

answer should not include h.
Physics
1 answer:
lions [1.4K]3 years ago
4 0

Answer:

t=\dfrac{d}{v_0cos(\theta )}

Explanation:

The background information:

<em>A student throws a water balloon with speed v0 from a height h = 1.8m at an angle θ = 29° above the horizontal toward a target on the ground. The target is located a horizontal distance d = 9.5 m from the student’s feet. Assume that the balloon moves without air resistance. Use a Cartesian coordinate system with the origin at the balloon's initial position.</em>

The time it takes for the balloon to reach the target is equal to the target distance d divided by the horizontal component of the velocity v_0:

t=\dfrac{d}{v_x}

where v_x is the horizontal component of the velocity v_0, and it is given by

v_x=v_0cos(\theta);

Therefore, we have

\boxed{t=\dfrac{d}{v_0cos(\theta)} }

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Answer:

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