Question

A ball is thrown into the air with an initial upward velocity of 48dt/s. Its height h in feet after t seconds is given by the equation h(t)=-16t^2+48t+4
A: What height will the ball be after 2 seconds?
B: After how many seconds will the ball reach its maximum haight?
C: What is the balls maximum height?

Answer 1

Answer:

Step-by-step explanation:

The equation used to represent the height of the ball, h in feet after t seconds is expressed as

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

A) The height of the ball after 2 seconds would be

h = - 16 × 2² + 48 × 2 + 4

h = - 64 + 96 + 4

h = 36 feet

B)The equation is a quadratic equation. The plot of this equation on a graph would give a parabola whose vertex would be equal to the maximum height travelled by the rocket.

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = -16

b = 48

Vertex = - - 48/32= 1.5

So the ball will attain maximum height at 1.5 seconds.

C) The maximum height of the ball would be

h = -16 × 1.5² + 48 × 1.5 + 4

h = - 36 + 72 + 4

h = 40 feet