SOLUTION: A metal ball (with a mass of 5.0 kg) is dropped from a building 320 meters tall.
How many seconds does it take to hit the ground?
What is the ball's velocity just before impact
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Question 971729: A metal ball (with a mass of 5.0 kg) is dropped from a building 320 meters tall.
How many seconds does it take to hit the ground?
What is the ball's velocity just before impact?
Thank you,
Madison
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A metal ball (with a mass of 5.0 kg) is dropped from a building 320 meters tall.
How many seconds does it take to hit the ground?
What is the ball's velocity just before impact?
=======================
The commonly used value for acceleration due to Earth's gravity is 9.8 m/sec/sec
------
Neglecting air friction:
--> h(t) = -4.9t^2 + 0t + 320 where t = seconds and h(t) = height AGL
At impact, h(t) = 0
h(t) = -4.9t^2 + 0t + 320 = 0
t^2 = 320/4.9
t =~ 8.081 seconds
==================
v = at = 9.8*8.081
=~ 79.196 m/sec
==========================
The mass is not a factor.
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
s= 1/2 at^2 a=9.8 m/sec units of s will be meters.
2s=at^2
t= sqrt ((2s/a)
t=8.08 sec.
vf=g*t
=9.8 m/sec^2 * 8 sec.
78.4 m/sec.
Note: if you set mgh=(1/2) mv^2, kinetic and potential energy, you get v= sqrt (2gh). This way gives 79.2 m/sec.
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