Question

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.

Answer 1

Answer:

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

Explanation:

The background information:

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.

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

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