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IceJOKER [234]
3 years ago
8

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

.1 m/s. After being thrown, the object falls freely due to gravity. Neglect air resistance and calculate the distance, in meters which the object covers between times t1 = 3.32 s and t2 = 5.08 s after it is thrown.
Physics
1 answer:
Softa [21]3 years ago
3 0

Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

t_1 = 3.32 sec

t_2 = 5.08 sec

from equation of motion we know that

d_1 = vt_1 + \frac{1}{2} gt_1^2

where d_1 is distance covered in time t1

sod_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2=

d_1 = 110.78 m

d_2 = vt_2 + \frac{1}{2} gt_2^2

where d_2 is distance covered in time t2

d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2

d_2 = 213.31 m

distance cover is  = 213.31 - 110.78 = 102.53 m

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