Physics, asked by amargupta1700, 6 months ago

A ball is dropped from the roof of a tower of height h. The total distance covered by it in the last
sec of its motion is equal to the distance covered by it in first three second. What is the value of
h? (g=10m/sec)​

Answers

Answered by gsadhana688
3

Answer:

the answer could be 30m.

Explanation:

h=g * t

or, h=10*3

or, h=30m

Answered by mrkamble70
2

Answer:

hey mate!!!!! here's your answer!!!!

Explanation:

A ball is dropped from the roof of a tower height= h

Total distance covered it in the last seconds of its motion...

= The distance covered it in first 3 seconds:

s= 1/2 at sq.

= 1/2 x 10 x 3 sq.

= 45

If the ball takes in second to fall to ground then the distance covered in nth second :

Sn= u + g/2 ( 2n-1)

= 0+ 10/2 (2xn-1)

=. 10 n - 5

Therefore, 45= 10n-5

= 10n = 50

Therefore, 1/2= 1/2 gt sq.

= 1/2 x 10 x 25

= 125m.

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