Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Answer:
<em> B.0</em>
Explanation:
Change in momentum: This is defined as the product of mass and change in velocity of a body. or it can be defined as the product of force and time of a body. The fundamental unit of change in momentum is kg.m/s
Change in momentum = M(V-U)......................... Equation 1
where M = mass of the ball, V = final velocity of the ball, U = initial velocity of the ball.
Let: M = m kg and V = U = v m/s
Substituting these values into equation 1
Change in momentum = m(v-v)
Change in momentum = m(0)
Change in momentum = 0 kg.m/s
<em>Therefore the momentum of the ball has not changed.</em>
<em>The right option is B.0</em>
