#### Answer

a. 8.9m/s
b. 10m

#### Work Step by Step

a. Use the level horizontal range formula to find the takeoff speed of v = 8.854 m/s. The range, takeoff angle, and g are known.
b. Let the end point be at y=0, and choose upward to be positive. The starting point is at y = 2.5 m, and the vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward.
Use equation 2.11b to solve for the time to fall a distance of 2.5 m below the starting height. The initial vertical velocity is $(8.854 m/s)(sin 45 ^{\circ})$. The time, which must be positive, is 1.597 seconds.
Now find the horizontal displacement during the jump.
$(8.854 m/s)(cos 45 ^{\circ})(1.597 s) = 10.0 m.$
She lands right where she is supposed to.