Question

plz solve the attachment
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Answer 1

Given:

Initial velocity,u, =40m/s

Final velocity,v, = 0

acceleration due to gravity ,g = -10m/s^2

When a body is thrown vertically upwards,it's velocity is decreasing, so the acceleration due to gravity,g= -10m/s^2

To Find:

• Height
• net displacement
• total distance covered

Solution:

1) Height ⬇️

${v}^{2} = {u}^{2} + 2gh \\ {(0)}^{2} = {(40)}^{2} + 2 \times ( - 10)h \\ 0 = 1600 - 20 \: h \\ h = \frac{1600}{20} \\ h = 80m$

Thus, the maximum height reached by the stone is 80 m.

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2) The stone is thrown up from the ground and after reaching the maximum height,falls back to the ground.As the final position of stone coincides with the initial position of stone , the net displacement of the stone is zero .

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3) The distance covered by the stone in reaching the maximum height is 80 metres. The stone will cover the same distance of 80 metres in coming back to ground.So, the total distance covered by the stone is 80+80 = 160 metres

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Answer 2

Given:

Initial velocity,u, =40m/s

Final velocity,v, = 0

acceleration due to gravity ,g = -10m/s^2

When a body is thrown vertically upwards,it's velocity is decreasing, so the acceleration due to gravity,g= -10m/s^2

To Find:

Height

net displacement

total distance covered

Solution:

1) Height ⬇️

$= > \: {u}^{2} = {u }^{2} + 2gh$

$= > ( {0})^{2} = {(46)}^{2} + 2 \times ( - 10)$

$= > 0 = 1600 - 20h$

$= > h = \frac{1600}{20}$

$= > h = 80m$

Thus, the maximum height reached by the stone is 80 m.

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2) The stone is thrown up from the ground

and after reaching the maximum height,falls back to the ground.As the final position of stone coincides with the initial position of stone , the net displacement of the stone is zero .

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3) The distance covered by the stone in reaching the maximum height is 80 metres. The stone will cover the same distance of 80 metres in coming back to ground.So, the total distance covered by the stone is 80+80 = 160 metres

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