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Problem 4:
A body fell vertically from the top of a tower. It covered 123.1 m in the final two seconds
before hitting the ground. Determine the height of the tower rounding your answer to the
nearest two decimal places. Let the acceleration due to gravity g=9.8 m/s2.

Respuesta :

Answer:

v2^2 - v1^2 = 2 g s        fundamental formula

v2 = v1 + 2 g    = v1 + 19.8         increase in velocity in 2 sec

v1^2 + 39.6 v1 + 392 - v1^2 = 2 * 9.8 * 123.1 = 2412.76

v1 = (2412.76 - 392) / 39.6 = 51.03

v2 = 51.03 + 19.6 = 70.63

T = 70.63 / .8 = 7.207 sec     time to fall height of tower

S = 1/2 g T^2 = 4.9 * 7.207^2 = 254.5 m

(Note  v2^2 - v1^2 = 70.63^2 - 51.03^2 = 2385 m

2385 / (2 * 9.8) = 122 m (close to 123.1    as was given

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