How to calculate time of a falling body with gravity?

Question

Say an object is falling in earth's gravational field and travels a distance of about 9.8 meters by which time its velocity would be about 13.8 m/second. Now imagine it hits an object below and its velocity is reduced to 8.9 m/second. How would you calculate the object's velocity if it fell another 9.8 meters if its velocity was now 8.9 m/second?

Update:

The title should really say velocity instead of time

denis · Accepted Answer

we can derive a basic kinematic equation using conservation of energy.  consider an object above the ground at a height of "ho".  Its total energy is simply the potential energy + kinetic energy provided there is an initial velocity "vo"

E = m*g*ho + ½*m*vo²

now let the object fall from a height "ho" to a final height "h".  now the object's total energy is composed of final kinetic energy and potential energy so we arrive at our wanted equation:

m*g*ho + ½*m*vo² = ½*m*v² + m*g*h

now we can simplify it:

g*ho + ½*vo² = ½*v² + g*h
v² = vo² - 2gΔh

where g = 9.8m/s² but we usually use down coordinate to be negative (g = -9.8m/s²) so we can rewrite as

v² = vo² + 2gΔh

now you are given values, just plug them in and solve for the final velocity

v² = 8.9² + 2(-9.8)(-9.8)
v = 16.47 m/s

Old Science Guy · Answer

...
You would have a new Vi (8.9)
Vf^2 = Vi^2 + 2 g d
Vf^2 = 8.9^2 + 2 (9.81) (9.8) = 271.5
Vf = 16.5 m/s

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How to calculate time of a falling body with gravity?

2 Answers