Question

A ball is projected vertically upward with a
speed of 50 m/s. Find
(i) the maximum height
(ii) the time to reach the maximum height,
(iii) the speed at half the maximum height.​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A ball is projected vertically upward with a speed of 50 m/s.

Find

(i) the maximum height

(ii) the time to reach the maximum height,

(iii) the speed at half the maximum height.

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Initial velocity (u) = 50 m/s
• Acceleration due to gravity (g) = - 10 m/s².

$\huge{\bold{\underline{Explanation:-}}}$

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Applying Third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

Here,

• Final velocity (v) = 0 m/s.
• Distance (s) =Maximum Height (H)

Substituting the values,

$\large{\tt \hookrightarrow (0)^2 - (50)^2 = 2 \times - 10 \times H}$

$\large{\tt \hookrightarrow 0 - 2500 = - 20 \times H}$

$\large{\tt \hookrightarrow - 2500 = - 20 \times H}$

$\large{\tt \hookrightarrow \cancel{-} 2500 = \cancel{-} 20 \times H}$

$\large{\tt \hookrightarrow 2500 = 20 \times H}$

$\large{\tt \hookrightarrow H = \dfrac{2500}{20}}$

$\large{\tt \hookrightarrow H = \cancel{\dfrac{2500}{20}}}$

$\huge{\boxed{\boxed{\tt H = 125 \: m}}}$

So, the Maximum height attained is 125 Meters.

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Applying First kinematics equation,

$\large{\boxed{\tt v = u + at}}$

$\large{\tt \hookrightarrow 0 = 50 + (- 10) \times t}$

$\large{\tt \hookrightarrow 0 = 50 - 10 \times t}$

$\large{\tt \hookrightarrow 10 \times t = 50}$

$\large{\tt \hookrightarrow 10t = 50}$

$\large{\tt \hookrightarrow t = \dfrac{50}{10}}$

$\large{\tt \hookrightarrow t = \cancel{\dfrac{50}{10}}}$

$\huge{\boxed{\boxed{\tt t = 5 \: Sec}}}$

So, the time taken by the body to reach Maximum height is 5 seconds.

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Now, At Half the Height.

⇒ h = H/2

⇒ h = 125/2 meters.

Applying Third kinematic equation,

$\large{\boxed{\tt v^2 - u^2 = 2as}}$

Here,

• s = h = 125/2 meters
• v ≠ 0 m/s.
• a = - g = - 10 m/s²

Substituting the values,

$\large{\tt \hookrightarrow v^2 - (50)^2 = 2 \times - 10 \times \dfrac{125}{2}}$

$\large{\tt \hookrightarrow v^2 - (50)^2 = \cancel{2} \times - 10 \times \dfrac{125}{\cancel{2}}}$

$\large{\tt \hookrightarrow v^2 - (50)^2 = - 10 \times 125}$

$\large{\tt \hookrightarrow v^2 - 2500 = - 1250}$

$\large{\tt \hookrightarrow v^2 = 2500 - 1250}$

$\large{\tt \hookrightarrow v^2 = 1250}$

$\large{\tt \hookrightarrow v = \sqrt{1250}}$

$\large{\tt \hookrightarrow v = 25 \sqrt{2}}$

√2 = 1.414

$\large{\tt \hookrightarrow v = 25 \times 1.414}$

$\huge{\boxed{\boxed{\tt v = 35.35 \: m/s}}}$

So, the Velocity at half of the maximum height is 35.35 m/s.

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