Answer:
Explanation:
We cannot give you the answer directly as we do not have her velocity calculated from Problem 1
Let's call that velocity v
when she jumps, her initial horizontal velocity is vcos45,
so the time needed to cross the horizontal distance is
t = d/v = 10/vcos45 seconds
The initial vertical velocity will be vsin45 m/s
now we can plug into the vertical analysis
h = h₀ + v₀t + ½at²
h = 2.5 + (vsin45)(10/vcos45) + ½(-9.81)(10/vcos45)²
which reduces to
h = 2.5 + 10tan45 - 4.905(100)/(v²cos²45)
h = 2.5 + 10(1) - 490.5/(0.5v²)
h = 12.5 - 981/v²
plugging in the initial leap speed, we get a value for h
If h is positive, it means she is still above the water line when she crosses the far shore line
If h is negative, it means she has splashed into the water before she crosses the far shore line.
setting h to zero, meaning she's at water level when she lands on the far shore, we can determine the critical launch speed.
0 = 12.5 - 981/v²
981/v² = 12.5
v² = 981/12.5
v = 8.86 m/s
If her launch velocity is greater than 8.86 m/s, she lands dry on the far shore.
If her launch speed is less than 8.86 m/s, she has a wet reception.
