Question

a body is thrown vertically upward with velocity of 98m/s find the max height attained by it the time taken to reach the starting -point

Answer 1

Given:-

• Initial velocity= 98m/s (u)
• Final velocity= 0m/s ( at maximum height velocity becomes 0, v)

To Find:-

• Height of body
• Time taken by body to reach at maximum height

_______________

Since, body goes against gravity so we take negative acceleration due to gravity (— g)

_______________

$\sf{ \sf{Using \: \: Newton's \: \: 1st \: \: law \: \: of \: \: gravitation}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \boxed{ \large\sf {\red {v = u + ( - g)t}}}}$

Now Substituting the given values in formula,

$\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies0 = 98 - 9.8t \: \: \: (since g = 9.8m {s}^{ - 2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies - 98 = - 9.8t$

$\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies9.8t = 98$

$\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies \: t = \frac{98}{9.8}$

$\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies \: \red{ t = 10s}$

_____________________

Now we calculate the height

$\sf {Using \: \: Newton's \: \: 2nd \: \: law \: \: of \: gravitation}$

$\large{ \boxed{ \red{ \sf{h = ut + \frac{1}{2} ( - g) {t}^{2} }}}}$

Now Substituting the given values in formula,

$\: \: \: \: \: \: \: \: \: \: \large \sf \implies h = 98 \times 10 + \frac{1}{2} \times ( - 9.8) \times {10}^{2}$

$\: \: \: \: \: \: \: \: \: \: \large \sf \implies h =980 + ( \frac{ - 49}{10} ) \times 100$

$\: \: \: \: \: \: \: \: \: \: \large \sf \implies h =980 - 490$

$\: \: \: \: \: \: \: \: \: \: \large \sf \implies \red{h =490m}$

Answer 2

Answer:

$\huge \pink{~} \red {A} \green {N} \blue {S} \orange {W} \pink {E} \red {R}$

Given:-

\begin{gathered} \\ \end{gathered}

\begin{gathered} \\ \end{gathered}

Initial velocity= 98m/s (u)

Final velocity= 0m/s ( at maximum height velocity becomes 0, v)

\begin{gathered} \\ \\ \end{gathered}

To Find:-

\begin{gathered} \\ \\ \end{gathered}

Height of body

Time taken by body to reach at maximum height

\begin{gathered} \\ \\ \end{gathered}

_______________

\begin{gathered} \\ \end{gathered}

Since, body goes against gravity so we take negative acceleration due to gravity (— g)

\begin{gathered} \\ \end{gathered}

_______________

\begin{gathered} \\ \end{gathered}

\sf{ \sf{Using \: \: Newton's \: \: 1st \: \: law \: \: of \: \: gravitation}}UsingNewton

′

s1stlawofgravitation

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { \boxed{ \large\sf {\red {v = u + ( - g)t}}}}

v=u+(−g)t

\begin{gathered} \\ \end{gathered}

Now Substituting the given values in formula,

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies0 = 98 - 9.8t \: \: \: (since g = 9.8m {s}^{ - 2}⟹0=98−9.8t(sinceg=9.8ms

−2

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies - 98 = - 9.8t⟹−98=−9.8t

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies9.8t = 98⟹9.8t=98

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies \: t = \frac{98}{9.8}⟹t=

9.8

98

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \: \: \large \sf \implies \: \red{ t = 10s}⟹t=10s

\begin{gathered} \\ \end{gathered}

_____________________

\begin{gathered} \\ \end{gathered}

Now we calculate the height

\sf {Using \: \: Newton's \: \: 2nd \: \: law \: \: of \: gravitation}UsingNewton

′

s2ndlawofgravitation

\begin{gathered} \\ \end{gathered}

\large{ \boxed{ \red{ \sf{h = ut + \frac{1}{2} ( - g) {t}^{2} }}}}

h=ut+

2

1

(−g)t

2

\begin{gathered} \\ \end{gathered}

Now Substituting the given values in formula,

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \large \sf \implies h = 98 \times 10 + \frac{1}{2} \times ( - 9.8) \times {10}^{2}⟹h=98×10+

2

1

×(−9.8)×10

2

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \large \sf \implies h =980 + ( \frac{ - 49}{10} ) \times 100⟹h=980+(

10

−49

)×100

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \large \sf \implies h =980 - 490⟹h=980−490

\begin{gathered} \\ \end{gathered}

\: \: \: \: \: \: \: \: \: \: \large \sf \implies \red{h =490m}⟹h=490m