Question #974b7

1 Answer
Oct 8, 2017

Answer:

See the answer below...

Explanation:

Here I am using a formula which you can check in my answer I replied before..

Here in this case there is acceleration i.e gravitational acceleration(#g=9.8ms^-2#) and the initial velocity i.e u is #0ms^-1# as it is only dropped...
Hence we can rewrite the equation as
#S=1/2g t^2#
#=>S=1/2xx9.8xx3.25^2=51.76m#

#color(red)(UPDATED#

Similarly there is a formula that #v^2=u^2+2as#
but here #v^2=2gs=>v=sqrt(2xx9.8xx51.76)=31.85ms^-1#

Hope it helps...
Thank you...