Question

A body is projected vertically upward. What is the distance covered in the last second of upward motion?
A) 2 m
B) 3 m
C) 4 m
D) 5 m
The initial velocity is not given so how can you find the answer? Please don't spam or you will be reported immediately.

Answer 1

Answer:

4.9 m or 5 m

Explanation:

t = 2u / g

= 2u / 9.8

t = u / 4.9

h = ut + 1/2 × g × t²

= u × u/4.9 + 1/2 × 9.8 × u/4.9 × u/4.9

= u²/4.9 + 4.9 × u²/24.01

= u²/4.9 + u²/4.9

= 2u² / 4.9

Now

v² - u² = 2gh

-u² = 2 × 9.8 × 2u²/4.9

-u² = 2 × 2 × 2u²

-u² = 8u²

2u² = 8

u² = 4

u = 2

Now,

Maximum height = u²/2g

= 2²/19.6

= 4/19.6 = 0.20m

But the options are 2m, 3m, 4m, 5m

So it should be reciprocal

= 19.6 / 4

= 4.9 m

Nearest value is 5 metres

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Answer 2

Answer:

Answer is given here:-

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