Question

A particle falls freely from the top of tower . If it covers the latter half of the distance in 10(√2-1)s. Then the time taken to reach the ground. If you answer this I will mark you as soon the brainliest . Please do as fast as possible.

Answer 1

Answer:

t2 = 10 * (2^1/2 - 1) = 4.14 s     time for second half of fall

h1 = 1/2 g t1^2         distance covered in first half of fall

h2 = v1 t2 + 1/2 g t2^2     distance covered in second half of fall

v1 = g t1

t1^2 = 2 t1 t2 + t2^2     setting h1 = h2

t1^2 - 2 t1 t2 - t2^2 = 0

t1^2 - 8.28 t1 - 17.1 = 0      substituting t2

t1 = 9.99 s  

T = t1 + t2 = 4.14 + 9.99 = 14.1 s     total time of fall

Check:

H = 1/2 g t^2 = 4.9 * 14.1^2 = 979 m

h1 = 1/2 g t1^2 = 4.9 * 9.99^2 = 489 m

h2 = 9.8 * 9.99 * 4.14 + 4.9 * 4.14^2 = 489 m