Answer to Question #720 in Mechanics | Relativity for Rooky
can you give answer, than show work?
y-y0= -1/2*g*t^2 + v_0*t
and since v_0=0 we get
y-y0= -1/2*g*t^2 => t=square-root[-2*(y-y0)/g]
we have y-y0= -1-0 = -1 m (we took the level of the table as y0=0 and the ground as y=-1m), g=9.81 m/s^2 so
The speed just before hitting can be deduced from
v-v_0= - g*t => v= -g*t+v_0 and since v_0=0 we get
v = -g*t = -9.81*0.45= 4.41 m/s
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