posted by Andy on .
A ball is thrown upward from a platform 5.1 meters high with a speed of 13 m/s at an angle 45 degrees from the horizontal. what is the magnitude of its velocity when it hits the ground?
No matter what the angle is, the ball will gain kinetic energy M g H, the potential energy decrease. The kinetic energy per mass is V^2/2.
Therefore V^2 will increase by 2gH, from 169 m^2/s^2 to 169 + 2*9.8*5.1 = 269 m^2/s^2.
The new velocity = 16.4 m/s