Question

A certain freely falling object, released from rest, requires 1.25 s to travel the last 22.5 m before it hits the ground.

(a) Find the velocity of the object when it is 22.5 m above the ground. (Indicate the direction with the sign of your answer. Let the positive direction be upward.)

(b) Find the total distance the object travels during the fall.

Answer 1

solution: Released from rest vi=0 Height released from = h. acceleration a=-9--9.81 m/s2 using s= 0; f+ 12 at² for last 1.255 - 22,5 - (V;) (1.25) -119.81) (1.25) velocity at the start of last 1.255 -22,5 +*19.81)(1.25) vis - = -11.87 m/s 1.25 (6) he object takes 1.25s to travel lost 22.5m. Total time of fall = t using s= Vit + 1/2 at? che 0 - gth and -(h-22.5) = g(+-1,25) h = 1/2gt² and h-22.5 = igit-1.25)? 1972– 22.5 = 19 (42 - 2.5++ 1.5625) -22,5 = 9(-2.57+1.5625) 1.5625+ z = 2.446 5 - 2.5 h = 1 at 2 h = 26.68 2 yr Total distance travelled during the fall = 26.68 m