(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
<u>The given parameters include:</u>
- constant velocity of the elevator, u₁ = 10 m/s
- initial velocity of the ball, u₂ = 20 m/s
- height of the boy above the elevator floor, h₁ = 2 m
- height of the elevator above the ground, h₂ = 28 m
To calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.

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Answer:
358.9 (+/- 0.4) million years ago
Holocene Epoch, of the Quaternary Period
Devonian period
66 million years ago (prox)
521 million years ago
110,000 years ago
NW
North america became more cold when it moved NW
Explanation:
Answer : The change in momentum of an object is equal to the impulse that acts on it.
Explanation :
Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.
The formula of change in momentum is,

Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of an object.

Proof :

Hence, the change in momentum of an object is equal to the impulse that acts on it.
Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x
N
modulus of elasticity E = 85 GN/m^2 = 85 x
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A =
=
= 1256.8 mm^2
area of hole a =
=
= 549.85 mm^2
Total contraction of the bar =
total contraction =
==>
= <em>1.228 x </em>
<em> mm </em>