Physics: Principles with Applications (7th Edition)

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.