From the question above, we know that Janet is moving horizontally at 2.33 m/s for 1.5 seconds.

2) How far from the base of the platform does she land? = 2.33 * 1.5 = 3.495 m

Since Janet is falling for 1.5 seconds, we can use the following equation to determine the distance she falls.

d = vi * t + ½ * 9.8 * t^2

vi initial vertical velocity = 0 m/s

d = ½ * 9.8 * 1.5^2 = 11.025 meters

This is the height of the platform.

1) Y=Yo+Vy(t) + (1/2)g(t²)===> Y=0 (its the water) Vy = 0 ( vertical velocity) g= (-9.8m/s²)

0 = Yo +0 +(1/2)(-9.8)(1.5²)

0 = Yo - 11.025 m

11.025m = Yo ( height of platform)

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2) Horizontal equations

X= Vx(t) where Vx = 2.33 m/s

X = 2.33(1.5) = 3.495 m