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A ball is thrown into the air with an upward velocity of 48 ft/s. Its height h in feet after

t seconds is given by the function h = -16t2 + 48t + 8. In how many seconds does the ball
reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height?

User GRS
by
5.5k points

2 Answers

0 votes

Answer:

1.5 seconds; 42 ft

Explanation:

User Ahmed Taher
by
4.2k points
5 votes

Answer:

Explanation:

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

h = -16t² + 48t + 8

The equation is a quadratic equation. The plot of this equation on a graph would form a parabola whose vertex would be equal to the maximum height travelled by the ball.

The vertex of the parabola is calculated as follows,

Vertex = -b/2a

From the equation,

a = -16

b = 48

Therefore,

Vertex = - 48/2 × - 16 = 48/32 = 1.5 seconds

So the ball will attain maximum height in 1.5 seconds.

To determine the maximum height, we would substitute t = 1.5 into the given equation. It becomes

h = -16(1.5)² + 48(1.5) + 8

h = - 36 + 72 + 8 = 44 feet

User Dan Breslau
by
4.7k points