Question

A stone is thrown up with a speed of 0.5ms-2 . How high will the stone go before it begins to fall?

Answer 1

Correct Question :

A stone is thrown vertically upward with a speed of 0.5m/s. How high will the stone go before it begins to fall ?

Solution :

Initial velocity = 0.5m/s

Acc. due to gravity = 10m/s²

For a body thrown vertically upward, g is taken negative.

Third equation of kinematics :-

• v² - u² = 2as

⇒ v² - u² = 2(-g)H

⇒ 0² - 0.5² = 2(-10)H

⇒ H = 0.25/20

⇒ H = 0.0125m

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Answer 2

Answer:

Question

A stone is thrown up with a speed of 0.5m/sec. How high will the stone go before it begins to fall?

Given

• initial velocity= 0.5m / sec
• final velocity= 0 m/sec
• gravity / acceleration= 10 m/sec

To Find

distance travelled

Concept

❖ When a ball is thrown vertically upward , it has some initial velocity and its travels the Maximum height till the final velocity becomes zero. The time taken to reach the highest point is called the time of ascent.

Solution

◕According to 3rd equation of motion:-

${ \underline{ \boxed{ \sf{ \large{ \color{grey}{ {v}^{2} = {u}^{2} + 2as}} }}}}$

where

• v denotes final velocity
• u denotes initial velocity
• s denotes Distance
• a denotes acceleration

Now subsitute the value in the above formula :-

$\sf { \to \:0 = {0.5}^{2} + 2 \times ( - 10) + \times s} \\ \\ \sf{ \to \: - 0.25 = - 20s} \\ \\ \sf{ \to \: s = \frac{0.25}{20}} \\ \\ \sf{ \to \: s =0.0125 \: metre }$

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