Help im confused

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

The initial vertical velocity will be vsin45

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 on the far shore.

If her launch speed is less than 8.86 m/s, she has a wet reception.

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered