Question

A body thrown vertically upwards with a speed of 19.6 ms–1 from the top of a tower returns to the earth in 10 seconds. What will be its height attained?

Answer 1

Answer:

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Explanation:

Given that,

Initial velocity u=19.6m/s

Final velocity v=0

Time t=6sec

The acceleration is

We know that,

v=u+at

0=19.6+a×6

a=−3.26m/s

2

Now, the height is

From equation of motion

s=ut+

2

1

at

2

s=19.6×6−

2

1

×3.26×36

s=58.9

s=59m

Hence, the height of the tower is 59 m

Answer 2

Answer:

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