Hello please reply to this question immediately

Asked by aarthi9 | 20th Jun, 2011, 02:09: AM

Expert Answer:

Let the height attained by the stone above the height of 5 mts is h and t1 is the time taken to attain it. t2 is the time taken to reach the ground from the highest point.  

For the motion from the height of 5 mts to highest point, using first equation of motion, we have

0 = u – gt1

u = gt1            ……………(1)

where u is the initial speed of the stone. Now using second equation of motion, we have

 

h = u t1 - ½ g t12

Using equ. (1),

h = gt1 t1 - ½ g t12 = ½ g t12       ………..(2)

 

For the motion from the highest point to the ground, using second equation of motion, we have

h + 5 = ½ g t22      (here the initial speed is zero)

 

eq. (2) – eq. (1)

 

5 = ½ g( t22 - t12)

(t2 + t1)( t2 - t1) = 1

(t2 - t1) = 1 / 3.2

(t2 + t1) = 3.2

2 t2 = 3.2 + 1/3.2 = (3.2 x 3.2 + 1) / 3.2

t2 = 1.76

Therefore, h = (½) x 10 x (1.76)2 – 5 = 10.49

Total distance covered = 2 h + 5 = 2 x 10.49 + 5 = 25.98 metre

 

Answered by  | 21st Jun, 2011, 01:19: PM

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