Question

An object thrown straight upward reaches a height of 10 meters. If the object is thrown upward with twice this speed, it will reach a height of ____ m.

Answer 1

Answer:

If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.

Explanation:

Based on the Principle of Energy Conservation we find that the square of the initial speed of the object ($v$), measured in meters per second, is directly proportional to its change in height ($\Delta h$), measured in meters. That is:

$v^{2} \propto \Delta h$

$v^{2} = k\cdot \Delta h$ (Eq. 1)

Where $k$ is the proportionality constant, measured in meters per square second.

Such constant is eliminated by using the following relationship:

$\left(\frac{v_{2}}{v_{1}} \right)^{2} = \frac{\Delta h_{2}}{\Delta h_{1}}$  

$\Delta h_{2} = \left(\frac{v_{2}}{v_{1}} \right)^{2}\cdot \Delta h_{1}$ (Eq. 2)

Where:

$v_{1}$, $v_{2}$ - Initial and final speed of the object, measured in meters per second.

$\Delta h_{1}$, $\Delta h_{2}$ - Initial and final changes in height, measured in meters.

If we know that $\frac{v_{2}}{v_{1}} = 2$ and $\Delta h_{1} = 10\,m$, then the change in height is:

$\Delta h_{2} = 2^{2}\cdot (10\,m)$

$\Delta h_{2} = 40\,m$

If the object is thrown upward with twice the initial speed, it will reach a height of 40 meters.

Answer 2

When the object is thrown with double the speed, it reaches a height of 40m.

Laws of motion:

Given that the height attained by the object is h = 10m.

Let the initial speed of the object be u

and its final speed will be v = 0 at the highest point.

Then from the third equation of motion, we get,

$v^2=u^2-2gh\\\\0=u^2-2gh\\\\h=\frac{u^2}{2g}$

if the object is thrown at double the speed, that is u' = 2u, then the height attained will be:

$h'=\frac{u'^2}{2g}\\\\h'=\frac{(2u)^2}{2g}\\\\h'=\frac{4u^2}{2g}\\\\h'=4h$

h' = 4 × 10m

h' = 40m

So, the height attained will be 40 m.

Learn more about laws of motion:

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