Question

Note that the answer is not:

A skydiver jumps off of an airplane. The plane flies at an altitude of 4 km, and at a speed of 100 km/hr. The skydiver "jumps" by simply letting go of the plane and falling out. Gravity is the only thing affecting their motion until they open their parachute at a height of 2.5 km. After opening their parachute, they experience constant acceleration until they land perfectly, with velocity when they hit the ground. Note, you may need to do some unit conversions for this problem. You can look up any that you don't know. Note, this problem ignores several effects on skydivers, like air resistance, so the numbers are not necessarily that realistic. V = 100 km/ur - 4 km 2.5km ty a,b) acceleration here=? tX horizontal c) distance ? b) What is the x component of the acceleration after the parachute opens, in meters per second squared? Use the coordinate system shown.

- 0 m/s^2

- 7.848 m/s^2

Answer 1

b> → For the vertical direction Ly-axis), From 4 km to 2.5 km of height, Vy ² - uy² => Vy² -(0) 2 9h (n = 4-2.5 = 1.5 km2 1500m] 249.81 m) 22) iSoom) 171.55 2 m/s (downward direction) Vy - from 2.5 km of height to the ground, (v'y) 2- (vyj 2 Qyſh') (0)– (171.552) 2 ay (-2500m) ау 5.886 m/q² ( upward direction) V'y Vy taj (-171.552) + (5.886) t => > t 29.146 €

→ For the horizontal direction (a-anis). From the height of 4 km to 2.5 km the horizontal relocity is 100 km / hr Una = 100 km/hr 27.778 mle 100 x 1000 m 3600 r From the height of 2.5 km to ground (which t = 29.416%), takes Vx = Ux + axit (27.778) + ax (29:416) ax 0.953 mlr (ore eign rignifies -x direction)