Question #720

A ball slowly rolls off a 1 meter high table and falls straight down. How long does it take for the ball to hit the florr and how fast will it be going just before it hits the floor? The acceleration of the ball due to gravity is 9.81 m/sec^2?
can you give answer, than show work?

Expert's answer

The time is given by the equation

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

t=square-root[2*(1)/9.81]=0.45 s

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

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

t=square-root[2*(1)/9.81]=0.45 s

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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