a-basketball-is-shot-upwards-from-the-ground-with-an-initial-velocity-of-40-and-

Question

A basketball is shot upwards from the ground with an initial velocity of 40 ft/s40	ext{ ft/s}. The height (in feet) of the basketball tt seconds after it is shot is given by the equation h(t)=−16t2+40th(t) = $$-16t^2 + 40t. $$What is the maximum height reached by the basketball?

Accepted Answer

25 ft25	ext{ ft}

Answer

20 ft20	ext{ ft}

Answer

35 ft35	ext{ ft}

Answer

40 ft40	ext{ ft}

A basketball is shot upwards from the ground with an initial velocity of 40 ft/s40\(\text{ ft/s}\). The height (in feet) of the basketball tt seconds after it is shot is given by the equation h(t)=−16t2+40th(t) = -16t^2 + 40t. What is the maximum height reached by the basketball?

A basketball is shot upwards from the ground with an initial velocity of $40 ft slash s$ . The height (in feet) of the basketball $t$ seconds after it is shot is given by the equation $h (t) = - 16 t^{2} + 40 t$ . What is the maximum height reached by the basketball?