determine-the-velocity-of-the-ball-after-ttt-seconds-which-is-launched-straight-

Question

Determine the velocity of the ball after tt seconds which is launched straight up from the top of a building with an initial speed of 40 ft/s40	ext{ ft/s} from a height of 60 ft60	ext{ ft} above the ground. The height (in feet) of the ball above the ground tt seconds after it is launched is given by the equation h(t)=−16t2+40t+60h(t)=$$-16t^2+40t+60.$$

Accepted Answer

v(t)=−32t+40v\left(t\right)=-32t+40

Answer

v(t)=−16t2+40v\left(t\right)=$$-16t^2+40$$

Answer

v(t)=−32t2+40v\left(t\right)=$$-32t^2+40$$

Answer

v(t)=−16t+40v\left(t\right)=-16t+40