Luke threw a ball into the air with an initial upward velocity of 48 ft/s. Its height, h, in feet after t seconds is given by the function h(t)=-16t^2+48t+4

a. what height will the ball be when 2 seconds have passed

b. in how many seconds with the ball reach its maximum height

c. what is the balls maximum height

d. after how many seconds will the ball hit the ground

Question

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Respuesta :

reeserl reeserl · Answer 1 · 2021-04-06T03:35:24+02:00

Answer:

a. 36 ft

b. Maximum height is reached in 1.5 sec.

c. Maximum height is 40 ft

d. 3.08 sec

Step-by-step explanation:

a. h = -16(2)^2+ 48(2) + 4

= -64 + 96 + 4 = 36 ft

b. Maximum height is at the vertex of the parabola.

Let (h, k) be the vertex.

h = -b/2a = - 48/-32 = 3/2 = 1.5

k = -16(1.5)^2 + 48(1.5) +4

-36 + 72 + 4 = 40

Vertex is (1.5, 40) 1.5 is the time it takes to reach maximum height and 40 is the maximum height

Maximum height is reached in 1.5 sec.

c. Maximum height is 40 ft

d. The 4 in the equation tells us that the ball was thrown initially from a height of 4 ft.

When the ball hits the ground, h(t) = 0

-16t^2+48t+4 = 0

-4(4t^2 - 12t - 1) = 0

Use the quadratic formula to solve for t

t = 3.08 sec