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.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...
