A 27.5 kg ball is thrown at an angle of 22.7° above the horizontal off of a building such that its initial potential energy is 423 J, and the initial kinetic energy is 347 J.

What will be the speed of the ball at the moment before it hits the ground?

*Remember to use the conservation of Energy!

7.48 m/s

Explanation:

Total energy of the ball doesn't change

So,

Initial energy = Final energy

Initial Kinetic + Initial Potential = Final Kinetic + Final Potential

347 + 423 = ( (1/2) * 27.5 * v^2 ) + ( 27.5 * 9.81 * 0 )

770 = 13.75 * v^2

v^2 = 56

v = 7.48

grav forces affect all objects magnetic forces affect magnetic objects and electric currents

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west