Question

a stone is released from the top of a tower of height 980m.calculate the velocity after 5seconds​

Answer 1

SOLUTION:-

Given

• Initial velocity (u) = 0

• height = 980m

• Time = 5 seconds

• g= 9.8 m/s²

To find

• velocity after 5 seconds?

Explanation

Using Formula ⤵️

v = u+gt

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• V= u+gt

Putting values ~

v= 0+9.8×5

v= 0+49

v= 49m/s

Therefore, the velocity obtained after 3 seconds is 49m/s

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Add for your knowledge ⤵️

☆ General equations of motion ☆

1) v = u+at

2) s = ut +½at²

3) v²= u²+2as

☆ Equations of motion for bodies thrown vertically upward ☆

1) v = u-gt

2) h= ut-½gt²

3) v²= u²-2gh

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Answer 2

★ QUESTION

A stone is released from the top of the tower at the height of 980 m. Calculate the velocity after 5 seconds.

GIVEN :

• Initial Velocity = 0

• Height of the tower = 980 m

• Time Taken to do the following activity = 5s

• Acceleration due to gravity = 9.8 m/s

TO FIND :

• The velocity after 5 seconds = ?

STEP - BY - STEP EXPLAINATION :

Now, applying the formula for finding velocity (v)

➠ Velocity (v) = u + gt

Substituting the values, as per the given formula :

➠ velocity (v) = u + gt

➠ velocity (v) = 0 + 9.8 × 5

➠ velocity (v) = 0 + 49

➠ velocity (v) = 49 m/s

Hence, we got here the velocity as

49 m/s

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⭐ADDITIONAL INFORMATION ⭐

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⟹ s = ut + ½at^2

⟹ v = u + at

⟹ v^2 = u^2 + 2as