How to solve for time till object hits ground when thrown upwards off a cliff?

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To calculate the time until an object hits the ground when thrown upwards off a cliff, use the equation d = vt + 1/2(at^2), where d is the height of the cliff (considered a negative value), v is the initial velocity, and a is the acceleration due to gravity (-9.8 m/s²). The object first moves upwards until its velocity reaches zero, then descends back down, passing the initial throw point. The total vertical displacement to consider is the height of the cliff, as you are ultimately solving for when the object reaches the ground. The quadratic formula can be utilized to find the time, as the upward and downward movements are accounted for in the displacement. This method effectively combines both the upward motion and the downward fall to determine the total time until impact.
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How do you properly solve for the time till an object hits the ground when its thrown upwards off a cliff? With only The initial velocity and the hieght of the cliff. Thx :smile:
 
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What goes up, must come down. It may not have been written but you were also given the acceleration due to gravity. (9.8m/s^2). Gravity is first acting against the object being thrown up, then it will reach a point where v = 0. Then the object will return down as gravity works with the object.
 
But this is going up from the top of the cliff, then down, passing the point where it was thrown. Do you have to do 2 separate calulations?
 
How do you properly solve for the time till an object hits the ground when its thrown upwards off a cliff? With only The initial velocity and the hieght of the cliff.

Find vertical info.
Vertical
d = vt + 1/2(atsquared)
you have v, d is the height of the cliff (a negative value) and you have a(-9.8m/ssquared which is gravity), solve for t using quadratic formula. The time vertically is the same as time horizontally but since you are only looking for time the horizontal info is not necessary.
 
lol but wut about the height that its thrown first, isn't it that plus the hieght of the cliff?
 
No because you are looking for the time when the ball reaches the ground. Therefore you have to ask yourself what will be the vertical displacement for the time you are looking for. It is -d. This makes sense because althought the ball will go up for a certain d, it will also come back down for that same d just in the opposite direction and will continue moving until it has fallen the height of the cliff (-d)
 

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