A batter hits a ball at initial velocity of 20m/s and its caught at same height. What velocity does the ball..? - i cathcher web
A batter hits a pop-up recorded directly on the colony at home with a top speed of 20 m / s. The Cather the same level was reached. How quickly the land the ball in the glove cathcher? (Neglect air resistance)
Sunday, January 31, 2010
I Cathcher Web A Batter Hits A Ball At Initial Velocity Of 20m/s And Its Caught At Same Height. What Velocity Does The Ball..?
3 comments:
This is a problem of energy conservation.
When the ball leaves the bat with a kinetic energy K = 1 mV / 2 ^ 2, which reached the maximum height H on the floor, was converted into potential energy mgh KE = PE. m is the mass of the ball, gg, and V = 20 m / s. It is, KE = 1 / 2 mv ^ 2 = mgh = PE.
When the ball reached, M, is accelerated again, all the way, how gravity works on the ball on the potential energy is converted into kinetic energy. Once the ball is the height h = 0, which is when he reaches the ball, there are more potential energy on the left side as compared to baseline. So, ta da, that all PE = mgh = 1 / 2 mv ^ 2 = KE become kinetic energy, again.
Thus, we set the two equations equal to each PE, we KE = 1 / 2 mv ^ 2 = PE = 1 / 2 mv ^ 2 = KE; shows that v = v In other words, the initial velocity V of the pole equal to velocity v end when the ball comes to bat. QED.
This is a problem of energy conservation.
When the ball leaves the bat with a kinetic energy K = 1 mV / 2 ^ 2, which reached the maximum height H on the floor, was converted into potential energy mgh KE = PE. m is the mass of the ball, gg, and V = 20 m / s. It is, KE = 1 / 2 mv ^ 2 = mgh = PE.
When the ball reached, M, is accelerated again, all the way, how gravity works on the ball on the potential energy is converted into kinetic energy. Once the ball is the height h = 0, which is when he reaches the ball, there are more potential energy on the left side as compared to baseline. So, ta da, that all PE = mgh = 1 / 2 mv ^ 2 = KE become kinetic energy, again.
Thus, we set the two equations equal to each PE, we KE = 1 / 2 mv ^ 2 = PE = 1 / 2 mv ^ 2 = KE; shows that v = v In other words, the initial velocity V of the pole equal to velocity v end when the ball comes to bat. QED.
This is a problem of energy conservation.
When the ball leaves the bat with a kinetic energy K = 1 mV / 2 ^ 2, which reached the maximum height H on the floor, was converted into potential energy mgh KE = PE. m is the mass of the ball, gg, and V = 20 m / s. It is, KE = 1 / 2 mv ^ 2 = mgh = PE.
When the ball reached, M, is accelerated again, all the way, how gravity works on the ball on the potential energy is converted into kinetic energy. Once the ball is the height h = 0, which is when he reaches the ball, there are more potential energy on the left side as compared to baseline. So, ta da, that all PE = mgh = 1 / 2 mv ^ 2 = KE become kinetic energy, again.
Thus, we set the two equations equal to each PE, we KE = 1 / 2 mv ^ 2 = PE = 1 / 2 mv ^ 2 = KE; shows that v = v In other words, the initial velocity V of the pole equal to velocity v end when the ball comes to bat. QED.
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