Determine the velocity of the ball after t t t seconds which is launched straight up from the top of a building with an initial speed of 40 ft/s 40\text{ ft/s} 40 ft/s from a height of 60 ft 60\text{ ft} 60 ft above the ground. The height (in feet) of the ball above the ground t t t seconds after it is launched is given by the equation h ( t ) = − 16 t 2 + 40 t + 60 h(t)=-16t^2+40t+60 h ( t ) = − 16 t 2 + 40 t + 60 .

v ( t ) = − 32 t + 40 v\left(t\right)=-32t+40

v ( t ) = − 16 t 2 + 40 v\left(t\right)=-16t^2+40

v ( t ) = − 32 t 2 + 40 v\left(t\right)=-32t^2+40

v ( t ) = − 16 t + 40 v\left(t\right)=-16t+40

Determine the velocity of the ball after ttt sec...

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