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Answer:53.63
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Explanation:
The equations of motion used in this question is 
When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.
But,due to gravity,the object accelerates downward at a rate of
.
In X-Direction,
Given that initial velocity=
=
Using
,

In Y-Direction,
Given that initial velocity=
=
Using
,



S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)
S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m
Distance double 720m*2=1440m
V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places
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Vo= 331+0.6T
360=331+0.6T
360-331=0.6T
29=0.6T
0.6T/29
T=6/290 so change it to simplest form and us formulas good luck