a body is dropped from a height of 320 M the acceleration due to the gravity is 10 M per second square (i) how long does it take to reach the ground? (ii) what is the velocity with which it will strike the ground?
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1) Using S=ut+1/2at^2
S=320(given)
u=0 (as it is dropped)
a=g= 10 (given)
320=0+1/2*10*t^2
320*2/10=t^2
t=8seconds Answer
2) Using v^2-u^2= 2aS
u=0
S=320
a=g= 10
v^2=2*10*320
v^2=6400
v=80m/s
Alternate
Usingv=u+at
u=0
a=g=10
t=8 (calculated above)
v = 10*8
v= 80m/s
S=320(given)
u=0 (as it is dropped)
a=g= 10 (given)
320=0+1/2*10*t^2
320*2/10=t^2
t=8seconds Answer
2) Using v^2-u^2= 2aS
u=0
S=320
a=g= 10
v^2=2*10*320
v^2=6400
v=80m/s
Alternate
Usingv=u+at
u=0
a=g=10
t=8 (calculated above)
v = 10*8
v= 80m/s
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