Question

A ball is thrown up with a speed of 15 meter per second, how high will it go before it begins to fall (g=10m/s²).
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Answer 1

Given :

• Final velocity = 15 m/s
• Gravitational force = 10 m/s²

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To Find :

• Distance Covered = ?

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Solution :

~ Formula Used :

• ${\underline{\boxed{\red{\sf{ {v}^{2} - {u}^{2} = 2gs }}}}}$

Where :

• ➬ v = Final velocity
• ➬ u = Initial velocity
• ➬ g = Gravitational force
• ➬ s = Distance Covered

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~ Calculating the Distance Covered :

${\dashrightarrow{\qquad{\sf{ {v}^{2} - {u}^{2} = 2gs }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ {15}^{2} - {0}^{2} = 2 \times 10 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 225 - 0 = 2 \times 10 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 225 = 2 \times 10 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ 225 = 20 \times s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \dfrac{225}{20} = s }}}} \\ \\ \\ \ {\dashrightarrow{\qquad{\sf{ \cancel\dfrac{225}{20} = s }}}} \\ \\ \\ \ {\qquad \; \; {\therefore \; {\underline{\boxed{\pmb{\color{darkblue}{\frak{ Distance \; Covered = 11.5 \; m }}}}}}}}$

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~ Therefore :

❛❛ The ball started to fall from the height of 11.5 metres . ❜❜

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