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A ball is launched upward at 14 m/s from a platform 30 m high.Find the maximum height the ball will reach and how long it will take.​

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Answer:

The ball will reach a maximum height of 39.993 meters after 1.428 seconds.

Explanation:

Let suppose that no non-conservative forces acts on the ball during its motion, then we can determine the maximum height reached by the Principle of Energy Conservation, which states that:


K_(1)+U_(g,1) = K_(2)+U_(g,2) (1)

Where:


K_(1),
K_(2) - Initial and final translational kinetic energies, measured in joules.


U_(g,1),
U_(g,2) - Initial and final gravitational potential energies, measured in joules.

By definition of translational kinetic energy and gravitational potential energy we expand and simplify the expression above:


(1)/(2)\cdot m\cdot v_(2)^(2)+m\cdot g\cdot y_(2)= (1)/(2)\cdot m\cdot v_(1)^(2)+m\cdot g\cdot y_(1) (2)

Where:


m - Mass of the ball, measured in kilograms.


g - Gravitational acceleration, measured in meters per square second.


v_(1),
v_(2) - Initial and final speed of the ball, measured in meters per second.


y_(1),
y_(2) - Initial and final heights of the ball, measured in meters.

The final height of the ball is determined by the following formula:


v_(2)^(2)+2\cdot g\cdot y_(2) = v_(1)^(2)+2\cdot g\cdot y_(1)


v_(1)^(2)-v_(2)^(2)+2\cdot g \cdot y_(1)=2\cdot g\cdot y_(2)


y_(2) = y_(1)+(v_(1)^(2)-v_(2)^(2))/(2\cdot g) (3)

If we know that
y_(1) = 30\,m,
v_(1) = 14\,(m)/(s),
v_(2) = 0\,(m)/(s) and
g = 9.807\,(m)/(s^(2)), the maximum height that the ball will reach is:


y_(2) = 30\,m + (\left(14\,(m)/(s) \right)^(2)-\left(0\,(m)/(s) \right)^(2))/(2\cdot \left(9.807\,(m)/(s^(2)) \right))


y_(2) = 39.993\,m

The ball will reach a maximum height of 39.993 meters.

Given the absence of non-conservative forces, the ball exhibits a free fall. The time needed for the ball to reach its maximum height is computed from the following kinematic formula:


t = (v_(2)-v_(1))/(-g) (4)

If we know that
v_(1) = 14\,(m)/(s),
v_(2) = 0\,(m)/(s) and
g = 9.807\,(m)/(s^(2)), then:


t = (0\,(m)/(s)-14\,(m)/(s) )/(-9.807\,(m)/(s^(2)) )


t = 1.428\,s

The ball will take 1.428 seconds to reach its maximum height.

User Sravan
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