Question

A ball is thrown straight up from a height of 3 ft with a speed of 32 ft/s. Its height above the ground after x seconds is given by the quadratic function y = -16x2 + 32x + 3.

Answer 1

Answer:

19ft

Step-by-step explanation:

Given the height of a ball above the ground after x seconds given by the quadratic function y = -16x2 + 32x + 3, we can find the maximum height reached by the ball since we are not told what to look for.

The velocity of the ball is zero at maximum height and it is expressed as:

V(x) = dy/dx

V(x) = -32x+32

Since v(x) = 0

0 = -32x+32

32x = 32

x = 32/32

x = 1s

Get the height y

Recall that y = -16x² + 32x + 3.

Substitute x = 1

y = -16(1)²+32(1)+3

y = -16+32+3

y = -16+35

y = 19ft

Hence the maximum height reached by the ball is 19ft