Question

A body was dropped from the top of a building at a height of 50 feet. Determine how many seconds it took to reach the ground.

Answer 1

liven height h=50 feet h=50430.48cm ha 1524 cm h - 15.2um speed when it dropped 15.24 m conservation of energy 1 m u 2 mgh v=2gh Us 2x9.8 x 15.24 v= 298.704 v= 17.28 m/sec Now using eqution of motion, U=utat ang 17,28 = 0 + 3.8 17.28 9.8 t= 1.76 sec

Answer 2

SOLUTION :

Distance covered by the object, s  = height of the  building = 50 ft

Initial velocity of the object, u = 0 ft/s

Final velocity when object reaches the ground = g t = 32.2 t ft/s

Average velocity during interval t = (0 + 32.2 t) / 2 = 16.1 t ft/s

So,

Distance covered, s = average velocity * t 

=> 50 = 16.1 t * t 

=> t^2 = 50 / 16.1 

=> t = sqrt(50 / 16.1)

=> t = time taken by the object to reach the ground = 1.76 sec. (ANSWER).