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Sunny_sXe [5.5K]
3 years ago
14

A marble is dropped off the top of a building. Find it's velocity and how far below to its dropping point for the first 10 secon

ds of the drop .Use g=-9.8m/s^2

Physics
1 answer:
WINSTONCH [101]3 years ago
4 0

1) velocity: 98 m/s downward

Explanation:

The marble moves by uniformly accelerated motion, with constant acceleration a=g=-9.8 m/s^2 directed towards the ground. Therefore, its velocity at time t is given by:

v(t)=v_0 +at

where

v_0 = 0 is the initial velocity

a=g=-9.8 m/s^2 is the acceleration

t is the time

Substituting t = 10 s, we find:

v(10 s)=0+(-9.8 m/s^2)(10 s)=-98 m/s

And the negative sign means the direction of the velocity is downward.


2) Distance covered: 490 m

The distance covered in an uniformly accelerated motion can be found with the formula:

S=\frac{1}{2}at^2

where

a=-9.8 m/s^2 is the acceleration

t is the time

Substituting t=10 s, we find

S=\frac{1}{2}at^2=\frac{1}{2}(-9.8 m/s^2)(10 s)^2=-490 m

And the negative sign means the displacement is below the dropping point.

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