A ball rolls off a table with a horizontal velocity of 4 m/s.?

Question

(a) If it takes 0.6 seconds for it to reach the floor, what is the vertical component of the ball's velocity just before it hits the floor? (Use g = 10 m/s2.)

(b) What is the horizontal component of the ball's velocity just before it hits the floor?

Answer 1

a)
v=at
10[m/s^2] * 0.6[s] = 6[m/s]
b)
There is no acceleration horizontally
4[m/s]

Answer 2

a) While rolling along the table, the ball has total energy TE(h) = KE + PE = 1/2 mv^2 + mgh; where h is the table height, m the ball's mass, and v = vx = 4 mps the rolling velocity of the ball in the horizontal direction. Just prior to impact with the ground, TE(0) = KE(x) + KE(y) as there is no PE when h = 0.

a) Conservation of energy says that TE(h) = 1/2 mvx^2 + mgh = !/2 mvx^2 + 1/2 mvy^2 = TE(0). The KE(x) terms cancel out leaving mgh = 1/2 mvy^2; so that vy^2 = 2gh and vy = sqrt(2gh) = sqrt(2g*1/2 gt^2) = gt ~ 6 meters/sec.

b) Assuming no work (WE) done by the ball from drag, then vx is constant = 4 mps.

Total velocity v^2 = vx^2 + vy^2 and the angle of impact is theta = arctan(vy/vx). Total KE = KE(x) + KE(y) = 1/2 m(vx^2 + vy^2) = 1/2 mv^2.

Answer 3

The horizontal velocity is beside the point. the only equation you decide on is "v=u + at" the place v is very final velocity, u is preliminary velocity (=0), a is acceleration(-9.eighty one m/s) and t is time (=0.5 s).

Answer 4

using g as 10m/s2 the final velocity will be 6m/s. The horizontal will 2.4m

Answer 5

d = .5gt^2
d = .5*10*(.6)^2
d = 5*(.36)
d = .72m

4 m/s