# Question #974b7

##### 1 Answer

Oct 8, 2017

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.euis#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...