Asked by JOY
A bullet is fired in a horizontal direction with a muzzle velocity of 300m\s in the absence of air resistance,how far will it have dropped in travelling a horizontal distance of (a)20m (b)40m (c)60m (d)how far will it drop in one second
Answers
Answered by
karen
In 20m that is 60s 40m is 7.5s 60m is 50s in one second is 300m
Answered by
Henry
a. Dx = Xo*t. t = Dx/Xo = 20/300 = 0.067 s.
Dy = 0.5g*t^2 = 4.9*0.067^2 = 0.022 m.
b. t = d/Vo = 40/300 = 0.133 s.
Dy = 4.9*0.133^2 = 0.087 m.
c. t = Dx/Vo = 80/300 = 0.267 s.
Dy = 4.9*0.267^2 = 0.348 m.
d. Dy = 4.9*1^2 = 4.9 m.
Dy = 0.5g*t^2 = 4.9*0.067^2 = 0.022 m.
b. t = d/Vo = 40/300 = 0.133 s.
Dy = 4.9*0.133^2 = 0.087 m.
c. t = Dx/Vo = 80/300 = 0.267 s.
Dy = 4.9*0.267^2 = 0.348 m.
d. Dy = 4.9*1^2 = 4.9 m.
Answered by
Nwankwo Nnaemeka
I don't understand
Please explain
Them
Please explain
Them
Answered by
Samuel
(a) S = ut+1/2at² (a=0)
S=ut
t= s/u
t= 20/300
t= 0.067 sec
h= ut+1/2gt² ( u=0)
h= 1/2*5*(0.067)²
h= 0.022 m
(b)s should be equal to 40m
(c) S = 60m
(d) h=ut+1/2gt² (u=0)
h=1/2gt² (g=9.8 or 10)
h= 1/2*9.8*1
h= 4.9m ( if g=10, h=5m)
S=ut
t= s/u
t= 20/300
t= 0.067 sec
h= ut+1/2gt² ( u=0)
h= 1/2*5*(0.067)²
h= 0.022 m
(b)s should be equal to 40m
(c) S = 60m
(d) h=ut+1/2gt² (u=0)
h=1/2gt² (g=9.8 or 10)
h= 1/2*9.8*1
h= 4.9m ( if g=10, h=5m)
Answered by
Anonymous
Tank u
Answered by
Anonymous
Explain better please
Answered by
Anonymous
Number b and c u did not explain them
Answered by
Kingsley
How do you get 4.9
Answered by
Anonymous
20×3=
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