Question

An object thrown vertically upwards reaches a height of 320 m. What was its initial velocity? (Assume g = 10 m/s2) *

100 m/s

0 m/s

80 m/s

90 m/s

​

Answer 1

Answer:

80 m/s

Explanation:

As per the provided information in he given question, we have :

• Final velocity (v) = 0 m/s [It has been thrown vertically upwards.]
• Height attained (h) = 320 m
• Acceleration due to gravity (g) = – 10 m/s² [Opposite direction to the direction in which gravitation acts]

Now, as we know that,

$\bigstar \quad \underline{\boxed{ \pmb{\frak{v^2 - u^2 = 2gh} } }}$

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height

$\longrightarrow \sf{\quad { (0)^2 - u^2 = 2 \times (-10) \times 320 }}$

$\longrightarrow \sf{\quad { - u^2 = -20 \times 320 }}$

$\longrightarrow \sf{\quad { - u^2 = -6400 }}$

Negative sign will be cancelled from both sides.

$\longrightarrow \sf{\quad { u^2 = 6400 }}$

$\longrightarrow \sf{\quad { u= \sqrt{6400} }}$

$\longrightarrow \quad \underline{\boxed{ \pmb{\frak{u =80 } }\; \pmb{\frak{m}} \: \pmb{\frak{s}}^{\pmb{\frak{-1}}} }}$

Therefore, initial velocity is 80 m/s.

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More Information :

Equations of motion for fréély falling bodies :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + gt \quad}} \\ \\ \pmb{\sf{ \quad \: h= ut}} + \cfrac{\pmb{\sf{ 1}}}{\pmb{\sf{ 2}}}\pmb{\sf{ g}}{\pmb{\sf{ t}}}^{\pmb{\sf{ 2}}} \quad \: \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2gh \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• g denotes acceleration due to gravity
• h denotes height
• t denotes time