A ball rolls off a table and falls 0.68 m to the floor, landing with a speed of 5.3 m/s. ?

Question

(a) What is the magnitude of the acceleration of the ball just before it strikes the ground?

(b) What was the initial speed of the ball?

(c) What initial speed must the ball have if it is to land with a speed of 6 m/s?

Answer 1

(a) a = v^2/2H; where v = 5.3 mps and H = .68 m

(b) u = 0 as the ball rolled off the table top

(c) v^2 = 36 = u^2 + 2aH = u^2 + (5.3)^2; so that u^2 = 36 - (5.3)^2

You can do the math.

The physics is this...there is a SUVAT equation v^2 = u^2 + 2aH that can be derived from the conservation of energy.  That is KE(0) = 1/2 mv^2 = 1/2 mu^2 + maH = KE(H) + PE(H), which says the kinetic energy just before impact KE(0) equals the kinetic energy at the top of the fall KE(H) plus the potential energy at that height PE(H).

Answer 2

Your working formula is

Vf^2 - Vo^2 = 2as

where

Vf = final velocity = velocity of ball as it hits the floor = 5.3 m/s (given)

Vo = initial velocity = 0 (as ball simply rolls off the table)

a = acceleration

s = distance that ball has to travel = 0.68 m (given)

<< What is the magnitude of the acceleration of the ball just before it strikes the ground? >>

Substitute appropriate values and solve for "a"

5.3^2 - 0 = 2(a)(0.68)

I trust that you can solve for "a" from the above.

<< What was the initial speed of the ball? >>

Initial speed of ball is equal to zero (since it simply rolled off the table).

<< What initial speed must the ball have if it is to land with a speed of 6 m/s? >>

Substitute

Vf = 6

g = 9.8

s = 0.68

in the above formula and solve for Vo.

Hope this helps.