Doubt in physics problems?

Question

1. an object falls freely from rest for 5 seconds. find the distance travelled in the last 2 seconds. (g = 9.8m/s^2)

2.A ball released from rest falls freely. it is found to fall through 53.9m in the 8th second of its journey. calculate the value of g.

Answer 1

1. u=0, t=5 seconds

using s=ut+(1/2)gt^2
distance travelled in 3 secs.
s3=(1/2)*9.8*3*3
=44.1 m

distance travelled in 5 sec
s5=(1/2)*9.8*5*5
=122.5 m

The distance travelled in last 2 secs.
=s5-s3
=122.5-44.1
=78.4 m

2.u=0
distance travelled in 7 secs
s7=(1/2)*g*49
similary
s8=(1/2)*g*64

Distance covered in 8th second=s8-s7
s8-s7=(1/2)*g*64-(1/2)*g*49
53.9=(1/2)*g*(64-49)
g=53.9*2/15
=7.187 m/sec^2

Answer 2

1.) s = (1/2)gt^2
s = (1/2)gT^2 - (1/2)gt^2 = 78.4m
2.) g = 2s/(T^2 - t^2) = 7.187 m/s^2