Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
Answer:
A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)
B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist
C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf
Say mass of husband is m1
Mass of the wife is m2
Velocity of the husband is v1
Velocity of the wife is v2
According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf
The momentum equation is
m1v1+m2v2= (m1+m2)vf
D. To solve for vf we need to make it subject of formula
vf= {(m1v1) +(m2v2)}/(m1+m2)
E. Substituting our given data
vf=
{(1570*58)+(2550*54)}/(1570+2558)
vf=91060+137700/4120
vf=228760/4120
vf=55.52m/s
Their speed after collision is 55.52m/s
The MA is 6! Hope This Helps!
Ideally the resistance should be ZERO
