Question

A player shoots a basketball from a height of 6 feet. The equation, h = -16t 2 + 25t + 6, gives the height, h , of the basketball after t seconds. Describe the height, rounded to the nearest tenth of a foot, of the basketball after 1.5 seconds, assuming no other player touches the ball.

Answer 1

Answer:

The height of the ball after t =1.5 second is 7.5 feet.

Description:

The basket ball shoots from a height of 6 feet, it increase the height 1.5 feet after 1.5 seconds.

Therefore, the basketball thrown at the rate of 1 feet/per second.

Step-by-step explanation:

Given: A player shoots a basketball from a height of 6 feet. The equation,

h(t) = $-16t^2 + 25t + 6$

We need to find the height of the ball when t = 1.5 and describe the height.

plug in t = 1.5 in h(t) = $-16t^2 + 25t + 6$

h(1.5) = $-16(1.5)^2 + 25(1.5) + 6$

Simplifying the above expression, we get

h(1.5) = -36 + 37.5 + 6

h(1.5) = -36+43.5

h(1.5) = 7.5 feet

The height of the ball after t =1.5 second is 7.5 feet

Description:

The basket ball shoots from a height of 6 feet, it increase the height 1.5 feet after 1.5 seconds.

Therefore, the basketball thrown at the rate of 1 feet/per second.

Answer 2

175t because the motion times the width gives you 175t