A particle is dropped from the top of tower .it covers 40 m in last 2 s .find height of tower
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Answered by
66
Let the body fall through the height of tower in t seconds. From,
D_n =u+a/2 (2n-1) we have,total distance travelled in last 2 seconds of fall is
D= D_t+D_((t-1) )
= [ 0+ g/2 (2t-1)]+ [0+g/2 {2(t-1)-1}]
=g/2 ( 2t-1)+g/2 (2t-3)= g/2 (4t-4)
= 10/2×4(t-1)
Or, 40 = 20(t-1) or t =2+1 = 3s
Distance travelled in t seconds is
s=ut +1/2 at^2 =0+1/2×10×3^2=45 m
so ans is 45m
D_n =u+a/2 (2n-1) we have,total distance travelled in last 2 seconds of fall is
D= D_t+D_((t-1) )
= [ 0+ g/2 (2t-1)]+ [0+g/2 {2(t-1)-1}]
=g/2 ( 2t-1)+g/2 (2t-3)= g/2 (4t-4)
= 10/2×4(t-1)
Or, 40 = 20(t-1) or t =2+1 = 3s
Distance travelled in t seconds is
s=ut +1/2 at^2 =0+1/2×10×3^2=45 m
so ans is 45m
Answered by
98
Use kinematical equation appropriatly. And you will gethe answert
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