solve the numerical based on ch 8 motion of class 9
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3.9 is time while 39 is velocity.
Distance is 78.4m and acceleration is 10m/s and u =0 put the values in equation s=ut+1/2at^2 and find t tgen put the values in v=u+at and find v
Distance is 78.4m and acceleration is 10m/s and u =0 put the values in equation s=ut+1/2at^2 and find t tgen put the values in v=u+at and find v
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konsa page no h bata do pls
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