A body falls freely under gravity (g = 9.8ms-2) from a height of 40m on to the top of a platform 0.8m above the ground. Its velocity on reaching the platform is
784ms-1
80ms-1
78.4ms-1
39.2ms-1
27.7ms-1
Explanation
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Using the motion formula:
v^2=u^2*2gs
where;
v=?
u=0(body falling from a height)
g=9.8m/s^2
s=40m-0.8m=39.2m
solution:
v^2=u^2*2as
v^2=2*9.8*39.2
since u=0
v^2=768.32
take the square root of both sides
v=square root of(768.32)
v=27.7ms^-1
v²=u²+2gs
thats the formula we use here
Parameters given are:
s(distance)=40m
=(40-0.8)m
=39.2m
the given value of "g"(acceleration due to gravity)=9.8m/s²
U(initial velocity)=0
v(final velocity)=unknown value (?)
*Solu_tion*:
v²=u²+2gs
substituting the values:
v²=0²+2×9.8×39.2
v²=768.32m/s
making "v" the subject of the formula:
take the square root of both sides to cancel out the square attached to "v":
√v²=√768.32m/s
v=27.7m/s. ✓✓ 
i hope this was helpful to you ...
kinetic energy after fall is equal to mgs, s= 40-0.8=39.2
mgs=1/2mv² cancle out mass you will have
gs= 1/2v²
9.8 * 39.2= 0.5v²
384.16/0.5= v²
768.32=V²
v= 27.7m/s
v² = u² +2as
v= ?
u=0
a=9.8
s= (40-0.8)= 39.2
v² = u² +2as
v²= 0+2×9.8×39.2
v²=768.32
v=√768.32
v= 27.7ms-¹.
