A0.500 kg ball is dropped from rest at a point 1.20 m above the floor. the ball rebounds straight upward to a height of 0.700 ma) what is the velocity one instant before the collision? b) what is the velocity one instant after the collision with the floor? c) what is the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

To solve this problem we will apply the concepts related to energy conservation. With this we will find the speed before the impact. Through the kinematic equations of linear motion we will find the velocity after the impact.

Since the momentum is given as the product between mass and velocity difference, we will proceed with the velocities found to calculate it.

Part A) Conservation of the energy

Part B)  Kinematic equation of linear motion,

Here

v= 0 Because at 1.5m reaches highest point, so v=0

Therefore the velocity after the collision with the floor is 3.7m/s

PART C) Total change of impulse is given as,

luster is not a good characteristic to use by itself because luster is the property of minerals that indicates how much the surface of a mineral reflects light. the luster of a mineral is affected by the brilliance of the light used to observe the mineral surface.

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.