Question

[tex]\Huge{\color{magenta}{\fbox{\textsf{\textbf{Question:-}}}}}
A ball is thrown vertically up. If the ball reached at maximum height in 3s. Assume air resistance is negligible. The maximum height of the ball is most nearly​

Answer 1

Given : Time taken to reach maximum height is 3s .

To Find : Find the maximum height reached

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SolutioN : For Solving this question we need to apply the first equation of motion . Let's Solve :

$\maltese$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ v = u + gt }}}}}$

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

Where :

• v = Final Velocity = 0 m/s
• u = Initial Velocity = ?
• g = Gravitational force = 10 m/s
• t = Time = 3s
• s = Maximum height = ?

$\maltese$ Calculating the Initial Velocity :

$\begin{gathered} \qquad \; \longrightarrow \; \sf { v = u + gt } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 0 = u + \bigg( - 10 \bigg) \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 0 = u + \bigg( - 30 \bigg) } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 0 = u - 30 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; \sf { 0 + 30 = u } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \longrightarrow \; {\underline{\boxed{\pmb{\sf{ Initial \; Velocity = 30 \; m/s }}}}} \; {\purple{\bigstar}} \\ \\ \\ \end{gathered}$

$\maltese$ Calculating the Maximum height :

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { {v}^{2} - {u}^{2} = 2gs } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { {0}^{2} - {30}^{2} = 2 \times \bigg( - 10 \bigg) \times s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { 0 - 900 = - 20 \times s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { - 900 = - 20 \times s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { \cancel{-} 900 = \cancel{-} 20 \times s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { 900 = 20 \times s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { \dfrac{900}{20} = s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \sf { \cancel\dfrac{900}{20} = s } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; {\underline{\boxed{\pmb{\sf{ Maximum \; Height = 45 \; m }}}}} \; {\red{\bigstar}} \\ \\ \\ \end{gathered}$

$\therefore \;$ Ball reached the height of 45 m .

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Answer 2

Given,

A ball thrown vertically up. Time taken to reach height is 3s.

Find the maximum height of the ball (s).

Solution :

Let's solve this problem by using Equations of motion,

according to the question,

♦ Final velocity (v) = 0m/s

♦ Initial Velocity (u) = ?

♦ Time (t) = 3s

♦ g = 10m/s

♦ Maximum Height (s) = ?

finding initial velocity (u) :-

by using,

$\ast \: \boxed{ \color{gold}\textbf{v = u + gt}}$

on substituting the values,

$\tt⇢ \: \: 0 = u + ( - 10)3$

$\tt⇢ \: \: u - 30 = 0$

$\tt⇢ \: \: u = 30m {s}^{ - 1}$

now, finding Maximum height reached by the ball in 3seconds :-

by using,

$\ast \: \boxed{ \color{gold} \bf {v}^{2} - {u}^{2} = 2gs}$

on substituting the values,

$\tt⇢ \: \: {0}^{2} - {30}^{2} = 2 \times - 10 \times s$

$\tt⇢ \: \: - 900 = - 20s$

$\tt⇢ \: \: s = \cancel{\frac{ - 900}{ - 20} }$

$\tt⇢ \: \: s = 45m$

FINAL ANSWER :

The maximum height reached by the ball in 3seconds is 45m .