How Long Does It Take for a Ball Launched Off a Cliff to Hit the Ground?

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To determine how long it takes for a ball launched off a cliff to hit the ground, the initial vertical velocity can be calculated using the equation V_y = V sin(θ). The time to reach the peak can be found by setting the final vertical velocity to zero and using the equation V_f = V_y + at, then doubling this time to find when the ball returns to the cliff level. For the descent to the ground, the final velocity can be calculated with V_f = √(V_i² + 2gh) and plugged into the appropriate kinematic equation. The complexity of the final result raised concerns about the correctness of the approach. Overall, the method used appears valid, but simplification may be necessary for clarity.
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Homework Statement



Ok, you have a cliff with height h and you launch a ball off of it at speed v and angle \theta. How long till it hit the ground?

Homework Equations



(1) Vsin(\theta) = V_{y}
(2) V_{f} = V_{y} + at
(3) d = ((V_{i} + V_{f})/2)*t

The Attempt at a Solution



Ok, so what I did was take equation (2) and set V_{f}= 0. This alllows me to solve for the time that it takes to reach peak. Multiply this result by 2 to get the time when the ball is back to the plane of the cliff. Then for the time down to the ground I used equation (3) and plugged in V_{f}= \sqrt{V^{2}_{i}+2gh} for V_{f}. Then I solved for t.

This lead to a result I feel looked to complicated for the question. Did I do this right?

Homework Statement


Homework Equations


The Attempt at a Solution

 
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Ok, I don't know why all my subscripts look like superscripts in this but they do. All those i's and f's are for intial and final velocity.
 
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