1.

If body travels half of its path in the last second of its fall from rest, find the time and height of its fall.

Answer» Let (T) be the total o fall of vertical distance (S). As per question distanc travelled in (T) the second is ` S//2` and distance travelled in time (T-1) cecond is S//2.
Here, u =0 , a= 9.8 ms^(-2) , D_T = S//2`
As ` D_t = u + a/2 (2 T-10, we have
`S//2 =0 + (9.8) /2 [2 T -1] = 4.9 ( 2T -1)` ...(i)
Roe distance travelled in ` (T -1) secinds, we use the relation,
` S= ut + 1/2 at^2`, we gat
`S/2 =0 + 1/2 xx 9.8 (T-1)^2 ` ..(iii)
Equating (i) and (ii), we have
` 4.9 ( 2 T-1) = 9.8 (T-1)^2`
or ` 2 T -1 = (T -10 ^2 = T^2 -2T + 1`
or `T^2 -4T + 2 =0`
:. ` T = (4 =- sqrt 16 -8)/2 = 2 +- sqrt 2 = (2 + 1.4140`
or ` (- 1.414 0 = 3. 414 s or 0. 586 S`
Time `0586 s` being less than ` 1 second` is not possible, since the total motion is for mote than ` 1 sectond`. Therefore ` T= 3.414 S`
Potting this value in (i) , we get
` S= 2 xx 4.9 (2 xx 3 .414 -1) = 57 .11 m`.


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