A rubber ball with a mass of 210 grams is thrown exactly horizontally from a height of 2 meters with an initial speed of 13 m/s. When it strikes the ground, it bounces off at the same angle it made when it landed. During the bounce, it loses 20% of it's kinetic energy to dissipative forces. How high does it go after bouncing?
I have this figured out but not sure if it is correct...if anyone can tell me how they would do it I would greatly appreciate it..thanks
2007-02-23 17:21:00 · 2 answers · asked by Kimmy D 1 in Science & Mathematics ➔ Physics
So you have figured out the "incident" velocity hitting the floor. It has two components: vertical and horizontal. Because the only forces are happening in the vertical direction (gravity and reaction from the floor), the horizontal momentum doesn't change so the horizontal velocity doesn't change. Only the vertical speed suffers a change (v(final vertical)^2 = 80% of v(initial vertical)^2 It ends up that the potential energy in the vertical direction is 80% of the original.
2007-02-23 17:33:24 · answer #1 · answered by Sir Richard 5 · 0⤊ 0⤋
let me give u hints...take gravity makes the ball accelerate 9.81 m/s ...ask ur teacher if he/she wants 9.81 m/s or 10 m/s then calculate...as for simple maths u definitely noe how to do rite?
2007-02-23 17:24:25 · answer #2 · answered by Blank 3 · 0⤊ 0⤋