Determine its speed immediately before landing

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The discussion focuses on calculating the speed of a 1000kg object just before it lands after falling from a 64m high cliff with an initial horizontal velocity of 100 m/s. The user correctly identifies that mass is irrelevant when ignoring air resistance and uses the SUVAT equations to analyze vertical motion, concluding that the time of descent is approximately 3.6 seconds. They calculate the vertical velocity just before landing as 36 m/s and maintain the horizontal velocity at 100 m/s. By applying vector addition, they arrive at a final speed of 106.3 m/s. The response confirms the validity of the approach and suggests that conservation of energy could also be used, though the current method is sound.
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Homework Statement



Please could someone check if I have got this right? Many thanks

The question: An object of mass of 1000kg leaves a 64m high cliff at 100 ms-1 before descending to the ground. Ignoring air resistance, determine its speed immediately before landing.


Homework Equations



I think its a projectiles question using the 'suvat' equations.

The Attempt at a Solution



If I'm ignoring air resistance then I assume that mass is irrelevant?

Vertically I have:

u = 0m/s
v = ?
a = 10m/s2
s = 64m
t = ?

Horizontally I have:

u = 100m/s
v = 100 m/s
a = 0m/s2
s = ?
t = ?


Using the vertical components:

s = ut + 1/2 at^2
so
64 = 0 + 1/2 x 10 x t^2
so
t^2 = (2x64)/10 = 12.8s
t = 3.6s

Then

v = u + at
= 0 + (10x3.6)
= 36m/s

That gives me a horizontal velocity of 100m/s and a vertical velocity of 36m/s which using vector addition (Pythagoras) gives an answer of 106.3m/s.

If someone could tell me if I've got that right that would be great, many thanks!



 
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Well it is best to use conservation of energy methods; the problem doesn't actually state that the object is projected horizontally off the cliff. But no matter, your solution using SUVAT looks good, since the speed will be the same regardless of the projection angle.
 
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