Answer:
[tex]t=\dfrac{d}{v_0cos(\theta )}[/tex]
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 [tex]d[/tex] divided by the horizontal component of the velocity [tex]v_0[/tex]:
[tex]t=\dfrac{d}{v_x}[/tex]
where [tex]v_x[/tex] is the horizontal component of the velocity [tex]v_0[/tex], and it is given by
[tex]v_x=v_0cos(\theta)[/tex];
Therefore, we have
[tex]\boxed{t=\dfrac{d}{v_0cos(\theta)} }[/tex]
