Question

A ball is tossed straight into the air from a bridge, and its height above the ground t seconds after it is thrown is f(t)=-16t^2+72t+56feet. A)How high above the ground is the bridge? B) Find the average rate of change of the ball over the 2 seconds. C) Find the velocity of the ball at 2 seconds. D) What is the maximum height the ball will reach? E) What is the velocity of the ball at the time when the ball is at its peak?

Answer 1

A)56 ft.
B)80 ft.
C) 68 ft/s
D)136ft.
E) 0 ft/s
not sure about b and c

Answer 2

The bridge is 56 ft above the ground, this is the constant of the function. The function states he threw the ball upwards at 72ft/s @56 ft above the ground

Answer 3

Use the difference quotient for part B,
\[\frac{f(2)-f(0)}{2}\]

Answer 4

For part c, take the first derivative of f(t) and plug in 2

Answer 5

Part D, find the vertex of the parabola

Answer 6

When the ball is at its peak, its velocity will be 0

Answer 7

its peak vertical velocity is 0 with the horizontal velocity constant i believe

Answer 8

Yeah but there is no horizontal velocity

Answer 9

there is no horizontal acceleration but there is a constant horizontal velocity right?

Answer 10

The ball is not moving horizontally at all, the problem states he threw the ball straight up

Answer 11

if he threw the ball straight up how could it form a parabola?

Answer 12

The graph of f(t) is a parabola

Answer 13

Not the trajectory

Answer 14

ooh okay i see now