Asked by susan
a bicyclist rides off a flat roof at 13.4m/s, the roof is 3.6m above the ground, how far from the edge of the building does the bicyclist land? round the answer to the nearest meter.
Answers
Answered by
Henry
h = Vo*t + 0.5g*t^2. = 3.6
0 * 4.9t^2 = 3.6
t^2 = 0.735 .
Tf = 0.857 s. = Fall time.
d = Xo * Tf = 13.4m/s * 0.857s. = 11 m.
0 * 4.9t^2 = 3.6
t^2 = 0.735 .
Tf = 0.857 s. = Fall time.
d = Xo * Tf = 13.4m/s * 0.857s. = 11 m.
Answered by
Yssah
Vo=13.4m/s
y=-3.6m (i really don't now if this should be negative, negative beacuse it says above the ground😅)
g=9.8m/s^2
0=means degree(if the problem didn't give a degree given it automatic zero)
Vertical and Horizontal Displacement Relationship Formula
y=xtan0 - gx^2/
2Vo^2cos^20
Formula
Vo=√gx^2 /
(xtan0-y)(2cos^20)
Substitute/Place the given
13.4m/s=√(9.8m/s^2) x^2 /
(xtan0- (-3.6m))(2cos^20)
Multiply
13.4m/s=√9.8m/s^2 x^2 /
(3.6m)(2)
Divide
13.4m/s=√9.8m/s^2 x^2 /
7.2m
Square root and squared (so square root and squared of x gone)
13.4m/s=√1.381111111 s^2 x^2
Divide both sides to left the x
13.4m/s / 1.166666667 s=1.166666667 s^2 x / 1.166666667 s
11.48571428 m = x
11 m or 11.49m = x
I hope you understand hehehe
y=-3.6m (i really don't now if this should be negative, negative beacuse it says above the ground😅)
g=9.8m/s^2
0=means degree(if the problem didn't give a degree given it automatic zero)
Vertical and Horizontal Displacement Relationship Formula
y=xtan0 - gx^2/
2Vo^2cos^20
Formula
Vo=√gx^2 /
(xtan0-y)(2cos^20)
Substitute/Place the given
13.4m/s=√(9.8m/s^2) x^2 /
(xtan0- (-3.6m))(2cos^20)
Multiply
13.4m/s=√9.8m/s^2 x^2 /
(3.6m)(2)
Divide
13.4m/s=√9.8m/s^2 x^2 /
7.2m
Square root and squared (so square root and squared of x gone)
13.4m/s=√1.381111111 s^2 x^2
Divide both sides to left the x
13.4m/s / 1.166666667 s=1.166666667 s^2 x / 1.166666667 s
11.48571428 m = x
11 m or 11.49m = x
I hope you understand hehehe
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