Question

A body is projected with velocity 9.8m/s vertically
upwards the distance travelled in last second of its
motion is (g =10m/s)
2) 10m
3) 35m
4) 45m
1) 5m​

Answer 1

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✰Question ✰

⋄ A body is projected with velocity 9.8m/s vertically upwards the distance travelled in last second of its motion is (g =10m/s)

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✰Given:-

⋄ Initial velocity of ball (u) = 9.8m/s

⋄ Final velocity (v) =0m/s

⋄ Acceleration due to gravity (g) =10m/s²

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✰ To Find :-

⋄ Distance (s) travelled by body.

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✰Solution :-

By using this formula

➧ v² =u² +2gh

➧ 0² = (10)² +2 × 9.8 ×h

➧ h = 100/19.6

➧h = 5.1 m (approx)

Answer 2

A body is projected with velocity 9.8m/s vertically  upwards the distance travelled in last second of its  motion is 5 m

Therefore, option (1) is correct.

Explanation:

Initial velocity $u=9.8$ m/s

Final velocity $v=0$

Let the time taken is t

Then, using the first equation of motion

$v=u-gt$

$0=9.8-10\times t$

$\implies t=\frac{9.8}{10}=0.98$ seconds

Distance travelled

using the third equation of motion

$v^2=u^2-2gh$

$0^2=9.8^2-2\times 10\times h$

$\implies h=\frac{9.8^2}{2\times 10}$

$\implies h=4.802$ m

or, $h=5$ m (approx.)

Since the body on,y takes approx 1 second to travel

Therefore, the distance travelled will be equal to the total distance covered by the body which is 5 m

Hope this answer is helpful.

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