a body is falling from a high minaret travels 40m in the last 2 second

1.	a body is falling from a high minaret travels 40m in the last 2 seconds of its fall to ground. height of minaret in meters is
Answer» ANSWER: This problem can EASILY be solved with the help of equation of motion. s = UT + 1/2 at^2 40 = 2U + 1/2.10.(2)^2 40 = 2u + 20 2u = 20 u = 10 m /s Now, v^2-u^2 = 2as 2.10.s=100 s = 5m Hence the total DISTANCE traveled is 45M.

a body is falling from a high minaret travels 40m in the last 2 seconds of its fall to ground. height of minaret in meters is

Answer»

ANSWER:

This problem can EASILY be solved with the help of equation of motion.

s = UT + 1/2 at^2

40 = 2U + 1/2.10.(2)^2

40 = 2u + 20

2u = 20

u = 10 m /s

Now,

v^2-u^2 = 2as

2.10.s=100

s = 5m

Hence the total DISTANCE traveled is 45M.

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a body is falling from a high minaret travels 40m in the last 2 seconds of its fall to ground. height of minaret in meters is​

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a body is falling from a high minaret travels 40m in the last 2 seconds of its fall to ground. height of minaret in meters is