Determine its speed immediately before landing

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SUMMARY

The problem involves calculating the speed of a 1000kg object that leaves a 64m high cliff at an initial horizontal velocity of 100 m/s, ignoring air resistance. The vertical motion is analyzed using the SUVAT equations, resulting in a time of 3.6 seconds to reach the ground and a vertical velocity of 36 m/s just before landing. The final speed is determined to be 106.3 m/s through vector addition of the horizontal and vertical components. The solution is confirmed as correct, and conservation of energy methods are suggested as an alternative approach.

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  • Understanding of SUVAT equations for motion analysis
  • Knowledge of vector addition for calculating resultant velocity
  • Familiarity with concepts of projectile motion
  • Basic principles of conservation of energy in physics
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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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