The Stratoshere tower is 1172 ft. high. It takes 1.3 min, 9 sto ascend from the

Question

The Stratoshere tower is 1172 ft. high. It takes 1.3 min, 9 sto ascend from the ground floor to the top of the tower using thehigh speed elevator. The acceleration of gravity is 9.81 m/s^2. Assumingthat the elevator starts and ends at rest, and its acceleration hasa constant magnitude when moving, find the acceleration of theelevator as a multiple of the acceleration of gravity. Answer inunits of g. The Stratoshere tower is 1172 ft. high. It takes 1.3 min, 9 sto ascend from the ground floor to the top of the tower using thehigh speed elevator. The acceleration of gravity is 9.81 m/s^2. Assumingthat the elevator starts and ends at rest, and its acceleration hasa constant magnitude when moving, find the acceleration of theelevator as a multiple of the acceleration of gravity. Answer inunits of g.

Explanation / Answer

The height of the stratoshere tower is H = 1172 ft = 1172 *0.3048 m = 357.22 m The time taken by the high speed elevator to ascend from theground floor to the top of the tower is t = 1.3 min,9s or t = ((1.3 * 60) + 9)s = 87s The acceleration of gravity is g = 9.81 m/s^2 Let the acceleration of the elevator be a.Therefore,weget H = ut + (1/2)(a + g)t2 Here,u = 0 m/s or 357.22 = 0 * 87 + (1/2) * (a + g) * (87)2 or (a + g) = (357.22 * 2/(87)2) or (a + g) = 0.094 or a = 0.094 - g = 0.094 - 9.8 = -9.706 m/s2 The negative value of acceleration indicates that the motionof the elevator is against the gravity. The negative value of acceleration indicates that the motionof the elevator is against the gravity.

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