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

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SUMMARY

The discussion focuses on calculating the time it takes for a ball launched off a cliff to hit the ground, given its height (h), initial speed (v), and launch angle (θ). The key equations utilized include Vsin(θ) = V_y for vertical velocity, V_f = V_y + at for final velocity, and d = ((V_i + V_f)/2)*t for distance. The solution involves setting the final vertical velocity (V_f) to zero to find the time to peak height, then using V_f = √(V_i² + 2gh) to determine the total time until the ball hits the ground. The approach is confirmed as valid despite initial concerns about complexity.

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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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