Question

NO IRRELEVANT ANSWERS.SPAM ANSWERS WILL BE REPORTED. A stone is thrown vertically upward with a velocity of 39.2m/s.find the velocity and height of the stone after second​

Answer 1

Answer:

Distance covered by freely falling stone at any instant t=1/2gt

2

Distance covered by upward thrown stone 19.6t−

2

1

gt

2

If they meet, sum=height of tower

2

1

gt

2

&19.6t−

2

1

gt

2

=39.9

19.6t=39.2

t=2s

And they would meet at height of 19.6t1/2gt

2

=39.2−19.6=19.6m

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Answer 2

Answer:

This is a simple question from NLM

Explanation:

Here is your answer

u=39.2 m/s

a=-9.8 m/s^2

t= 1 s ( I am taking 1 since you have not specified , if something else is given then refer to this answer for help )

By first equation of motion

v=u+at

v=39.2+(-9.8)×1

v=39.2-9.8

v=29.4 m/s

By 2nd equation of motion

s=ut +(at^2)/2

s= 39.2×1 + (-9.8)×1×1

s=39.2 -9.8

s= 29.4 m

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