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.

Answer 1

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.

Answer 2

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