Question

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.​

Answer 1

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.