You throw a rock straight up into the air with a speed of 14.2 m/s. how long does it take the rock to reach its highest point? A. 2.1 s
B. 1.4 s
C. 0.7 s
D. 1.2 s


Answer 1
The acceleration of gravity on or near the Earth's surface is 9.8 m/s² downward.
Is that right ?           I don't hear any objection, so I'll assume that it is.

That means that during every second that gravity is the only force on an object,
the object either gains 9.8m/s of downward speed, or it loses 9.8m/s of upward
speed.   (The same thing.)

If the rock starts out going up at 14.2 m/s, and loses 9.8 m/s of upward speed
every second, it runs out of upward gas in (14.2/9.8) = 1.449 seconds (rounded)

At that point, since it has no more upward speed, it can't go any higher.  Right ?

(crickets . . .)

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A 25,000-kg train car moving at 2.50 m/s collides with and connects to a train car of equal mass moving in the same direction at 1.00 m/s. How much does the kinetic energy of the system decrease during the collision?



14062.5 J


From the law of conservation of momentum,

Total momentum before collision  = Total momentum after collision.

V = (m₁u₁ + m₂u₂)/(m₁+m₂).................1

Where V = common velocity after collision

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 1

V = [25000(2.5) + 25000(1)]/(25000+25000)

V = (62500+25000)/50000

V = 87500/50000

V = 1.75 m/s.

Note: The collision is an inelastic collision as such there is lost in kinetic energy of the system.

Total Kinetic energy before collision = kinetic energy of the first train car + kinetic energy of the second train car

E₁ = 1/2m₁u₁² + 1/2m₂u₂²........................ Equation 2

Where E₁ = Total kinetic energy of the body before collision, m₁ and m₂ = mass of the first train car and second train car respectively. u₁ and u₂ = initial velocity of the first train car and second train car respectively.

Given: m₁ = m₂ = 25000 kg, u₁ = 2.5 m/s, u₂ = 1 m/s

Substitute into equation 2

E₁ = 1/2(25000)(2.5)² + 1/2(25000)(1.0)²

E₁ = 12500(6.25) + 12500

E₁ = 78125+12500

E₁ = 90625 J.


E₂ = 1/2V²(m₁+m₂)....................... Equation 3

Where E₂ = total kinetic energy of the system after collision, V = common velocity, m₁ and m₂ = mass of the first and second train car respectively.

Given: V = 1.75 m/s, m₁ = m₂ = 25000 kg

Substitute into equation 3

E₂ = 1/2(1.75)²(25000+25000)

E₂ = 1/2(3.0625)(50000)

E₂ = (3.0625)(25000)

E₂ = 76562.5 J.

Lost in kinetic Energy of the system = E₁ - E₂ = 90625 - 76562.5

Lost in kinetic energy of the system = 14062.5 J

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