Question

A ball is dropped from a window 20 metres
high. How long will it take to reach the
ground? (g=10m/s)​

Answer 1

Answer:

2 sec

Explanation:

we know,

we know,s=h=20

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2as

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/s

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/sfor calculating time

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/sfor calculating timewe know,

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/sfor calculating timewe know,v=u+at

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/sfor calculating timewe know,v=u+at20= 0+ 10xt

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/sfor calculating timewe know,v=u+at20= 0+ 10xtt=2seconds

we know,s=h=20a=g=±10m/s=10m/s (since it is dropped)v^2-u^2= 2asv^2=2x10x20 (u=0; it is dropped not thrown)v=20m/sfor calculating timewe know,v=u+at20= 0+ 10xtt=2secondsthus, it will attain a speed of 20m/s before touching the ground, but the time taken to reach the ground is 2seconds. Because the velocity is not uniform but acceleration/gravity is. If still there is any confusion. Try the same problem with 80 of height.

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