Abaseball player hit a ball with an upward velocity of 46 ft/s. its height in feet after t seconds is given by the function h= -16t^2+46t+6. what is the maximum height the ball reaches? how long will it take to reach maximum height? how long does it take to hit the ground?

Given the function h(t) where t is the time, to find the maximum height reached by the ball, we must find the maximum value of h(t) = -16t² + 46t + 6. Since h(t) is a quadratic function, to find its maximum value, we need to look for the coordinates of its vertex (x, y). To solve for the x-coordinate, we have




Thus, it takes t = 23/16 s for the ball to reach the maximum height. To find the maximum height, we need to substitute the value of t into h(t). 

h = -16(23/16)² + 46(23/16) + 6 
h = 625/16 ft

For the ball to reach ground level, then h must be equal to zero. Thus, we have

0 = -16t² + 46t + 6 
0 = 4t² - 23t - 3

Recalling the quadratic formula, we have



where a , b, and c are the coefficients of the quadratic equation. For our case, a = 4, b = -23, and c = -3. Thus, we have




Since time cannot be negative, we disregard the negative root and have t = (23 + √897)/8 s.

 
Maximum time = 23/16 seconds
Maximum height = 625/16 ft
Time taken to reach the ground = (23 + √897)/8 seconds
 

The answer is b) 32 your welcome