If a mass of 4 kg is dropped from da height of 10m,what will be its velocity when it reaches the ground level ?(g=9.8 m/sec 2)
Velocity of body when reaches da ground level=?
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Mass= 4Kg
Height=10m(let it be 's')
u (initial velocity)= 0m/s
acceleration due to gravity = 9.8m/s^2
According to laws of motion,
v^2-u^2= 2as
=> (v)^2- 0= 2*(9.8)*(10)
=(v)^2=2x98 {9.8x10=98}
v^2= 196
=>v = sqrt of 196 = 14m/s
Final velocity = 14m/s
Height=10m(let it be 's')
u (initial velocity)= 0m/s
acceleration due to gravity = 9.8m/s^2
According to laws of motion,
v^2-u^2= 2as
=> (v)^2- 0= 2*(9.8)*(10)
=(v)^2=2x98 {9.8x10=98}
v^2= 196
=>v = sqrt of 196 = 14m/s
Final velocity = 14m/s
Monty1992:
There r 4 option
34.27m/sec,40.27m/sec,44.27 m/sec ,50.27m/sec
sorry? i didn't get you... ):p
There are 4 options.
Answered by
0
according to Newton's third equation
v²-u²=2aS
v²-0=2*9.8*10
v²=196
v=14m/s
so the final velocity was 14 m/s
v²-u²=2aS
v²-0=2*9.8*10
v²=196
v=14m/s
so the final velocity was 14 m/s
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