What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. - air methods englewood
A ball is thrown from the roof of a building height of 21.6 m with an initial speed of 11.7 m / s and at an angle of 57.5 degrees above horizontal.What the speed of the ball just before it touches the ground ? The distribution of energy consumption and ignore air resistance.
Take advantage of the free fall acceleration to be = 9.80 m / s 2 ^
Friday, January 1, 2010
Air Methods Englewood What Is The Speed Of The Ball Just Before It Strikes The Ground? Use Energy Methods And Ignore Air Resistance.
2 comments:
Because the ball is out of the roof of a building height of 21.6 m with an initial speed of 11.7 m in height thrown / s, baseball has a kinetic energy (KE) and potential energy (PE)
Initial KE = (1 / 2) mu ^ 2 = (1 / 2) m * 11.7 * 11.7 = 68,445 J m
First PT = mgh = m * 9.8 * 21.6 = 211.68 m J
First total energy E = KE + PE
First total energy E = m + 68,445 m = 211.68 m 280.125 J
First total energy E = 280.125 J m
If the ball touches the ground, which is not final and final PE KE (1 / 2) mv ^ 2, where V is the velocity when it hits the ground.
Final total energy E! = KE final + PE final
Final total energy E! = (1 / 2) mv ^ 2 + zero
Neglecting the air resistance is not received, the total energy
Final total energy E! First total energy E =
(1 / 2) mv ^ 2 = 280,125 J m
m Style
V ^ 2 = 2 x 280,125 = 560.25
V = 560.25 sq rt = 23.67 m / s
The speed of the ball just before it touches the ground, is 23.67 m / s
Because the ball is out of the roof of a building height of 21.6 m with an initial speed of 11.7 m in height thrown / s, baseball has a kinetic energy (KE) and potential energy (PE)
Initial KE = (1 / 2) mu ^ 2 = (1 / 2) m * 11.7 * 11.7 = 68,445 J m
First PT = mgh = m * 9.8 * 21.6 = 211.68 m J
First total energy E = KE + PE
First total energy E = m + 68,445 m = 211.68 m 280.125 J
First total energy E = 280.125 J m
If the ball touches the ground, which is not final and final PE KE (1 / 2) mv ^ 2, where V is the velocity when it hits the ground.
Final total energy E! = KE final + PE final
Final total energy E! = (1 / 2) mv ^ 2 + zero
Neglecting the air resistance is not received, the total energy
Final total energy E! First total energy E =
(1 / 2) mv ^ 2 = 280,125 J m
m Style
V ^ 2 = 2 x 280,125 = 560.25
V = 560.25 sq rt = 23.67 m / s
The speed of the ball just before it touches the ground, is 23.67 m / s
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