Find final velocity in a horizontal projectile motion

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To find the final velocity in horizontal projectile motion, the height of the cliff can be calculated using the equation Δy = 2⋅g⋅t^2, resulting in a height of 30.6 meters. The vertical final velocity (Vfy) is determined to be 24.5 m/s after 2.5 seconds. However, the horizontal distance (Δx) cannot be calculated without knowing the initial horizontal velocity (Vx). If the angle of projection is 0°, the initial velocity does not affect the time to hit the ground, and the final horizontal velocity remains constant. Without additional information on initial velocity, the horizontal distance cannot be accurately determined.
vAhmed
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Homework Statement


A ball was thrown from a cliff, it reached to the ground after 2.5 seconds.
a) Find the height of the cliff.
b) Find the horizontal distance the ball reached.

Homework Equations


Δy = 2⋅g⋅t^2
Vfy = g⋅t
√(Vfy)^2+(Vfx)^2
Δx= Vx⋅t

The Attempt at a Solution


I've found Δy=30.6m and Vfy=24.5m/s but I couldn't find Vfx without Δx using the equation Δx=Vx⋅t
How can I find Vfx in order to calculate Vf
 
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You need to know the initial velocity and the angle which thrown. If it was thrown at a 0° angle to the horizontal, then the initial velocity will not contribute to how long it takes to hit the ground.

If you are assuming negligible air resistance, then you can assume the final horizontal velocity will be approximately equal to the initial horizontal velocity.

If the problem doesn't give an initial velocity, You could take something known like the fastest baseball thrown, and use that to find an upper bound.o0)
 
+1

vAhmed said:
I've found Δy=30.6m and Vfy=24.5m/s

Those answers look ok but as Scott said you don't have enough info in the problem statement to answer b). Did you miss out something?
 
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