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

Question

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

Accepted Answer

$v (t) = - 32 t + 40$

Answer

$v (t) = - 32 t + 40$

Answer

$v (t) = - 32 t + 40$

Answer

$v (t) = - 32 t + 40$