Question

At a friend

Answer 1

during the trajectory the horizontal speed of the quarter doesn't change and is given by 5.3 cos 45

horizontal speed = 5.3 cos45 = 3.75 m/s

so time taken by the quarter to reach the dish = horizontal distance / horizontal speed = 2.1 / 2.78 = 0.56 sec

the initial vertical speed of the quarter = 5.3 sin45 = 3.75 m/s

vertical accleration of the quarter = -g = -9.8m/s^2

so the height of the shelf = vertical displacement of the quarter = ut - 0.5gt^2 = 3.75*0.56 - (0.5 * 9.8 * 0.56*0.56) = 0.56m

so height of the shelf = 0.56m

vertical velocity of quarter while landing in the dish = u - at = 3.75 - 9.8*0.56 = -1.74

final velocity = sqrt [ (horizontal velocity)^2 + (vertical velocity)^2] = sqrt( 3.75^2 + (-1.74^2) ) = 4.13 m/s

direction = arctan (vertical velocity / horizontal velocity) = arctan (-1.74/3.75) = 25 degree below horizontal

so final velocity = 4.13 m/s at 25 degrees below horizontal