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torisob [31]
3 years ago
14

An object whose mass is 100 lb falls freely under the influence of gravity from an initial elevation of 600 ft above the surface

of Earth. The initial velocity is downward with a magnitude of 50 ft/s. The effect of air resistance is negligible. Determine the velocity, in ft/s, of the object just before it strikes Earth. Assume g = 31.5 ft/s
Physics
1 answer:
monitta3 years ago
7 0

Answer with Explanations:

Given:

Mass of object, m = 100 lb

height fallen, h = 600 ft

initial velocity, u = 50 ft/s

acceleration due to gravity, g = 31.5 ft/s^2

Find final velocity when it touches ground.

Solution:

Use standard kinematics equation, in the absence of air resistance and variation of g with height,

v^2 - u^2 = 2aS

where

v = final velocity

u = initial velocity

a = acceleration due to gravity

S = distance travelled

Substitute values

v^2 = u^2 + 2aS

= 50^2 + 2*31.5*600

= 40300 ft^2/s^2

Final velocity,

v = sqrt(40300) ft/s

= 200.75 ft/s

= 201 ft/s  to the nearest foot.

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A 1459 kg car is traveling WEST at 43 m/s. A 9755 kg truck is traveling EAST at 11 m/s. They collide head-on, and stick together
otez555 [7]

Answer:

<em>Both vehicles move east at 3.97 m/s</em>

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<u>Law Of Conservation Of Linear Momentum </u>

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is:

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

P=m_1v_1+m_2v_2

If a collision occurs and the velocities change to v', the final momentum is:

P'=m_1v'_1+m_2v'_2

Since the total momentum is conserved, then:

P = P'

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

Assume both masses stick together after the collision at a common speed v', then:

m_1v_1+m_2v_2=(m_1+m_2)v'

The common velocity after this situation is:

\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Assuming east direction to be positive, we have an m1=1459 kg car traveling west at v1=-43 m/s. An m2=9755 kg truck is traveling east at v2=11 m/s. They collide head-on and stick together after that.

Computing the resultant velocity after the collision:

\displaystyle v'=\frac{1459*(-43)+9755*11}{1459+9755}

\displaystyle v'=\frac{44568}{11214}

v' = 3.97 m/s

Both vehicles move east at 3.97 m/s

4 0
3 years ago
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