A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (?Fig. E2.44?). After the sandbag is released, it is in free fall. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release does the bag strike the ground? (c) With what magnitude of velocity does it strike the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch ay-t, vy-t, and y-t graphs for the motion.

Solution 44E Consider the formulas for position and velocity 2 x = x0+ v t0+ 1/2 at v = v0+ at Where x = final position x = original position 0 v0= original velocity a = acceleration v = final velocity (a). x = x + v t + 1/2 at2 0 0 2 2 x = 40.0 m + 5.0 m/s(0.250 s) + (0.5)( 9.8 m/s )(0.250 s) x = 41.556 m. (b). v = v + at 0 v = 5.0 m/s + ( 9.8 m/s )(0.250 s) v = 2.55 m/s. 2 (c).x = x 0 v t0+ 1/2 at x = 40.0 m + 5.0 m/s(1.00 s) + (0.5)( 9.8 m/s )(1.00s ) 2 x = 49.9 m. (d).v = v 0 at v = 5.0 m/s + ( 9.8 m/s )(1.00 s) v = 4.8m/s. (e).x = 0 to solve for time. 2 0 = 40 + 5(t) + 1/2( 9.8)t Using quadratic formula. t = b ± 4ac/2a Here a = 9.8 a = 5 a = 40 t = 9.166 s Using this time put into velocity formula v = v + at 0 v = 5 + ( 9.8)(9.166) v = 84.83 s To find greatest height,we need to find the time it takes for velocity to reach 0 m/s. So we can take final velocity as zero. 0 = 5 + ( 9.8)t t = 0.510204 s Put this time into position formula x = x + v t + 1/2 at 2 0 0 x = 40.0 m + 5.0 m/s(0.510204 s) + (0.5)( 9.8 m/s )(0.510204 s) 2 x = 41.2775 m Time v/s velocity graph is shown below. Y v/s t graph is as shown below.