Me, i'm trying to !
a ball is thrown from a height of 151 feet with an initial downward velocity of 15/fts . the ball's height h (in feet) after t seconds is given by the following.

h=151−15t-16t^2

how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth.

2.64 seconds

Step-by-step explanation:

To find when the ball will hit the ground, you just need to use the equation of h with the value h = 0 (the ball is in the ground), and then find the value of t:

h = 151 − 15t - 16t^2

0 = 151 − 15t - 16t^2

To solve this quadratic equation, you can use Bhaskara formula:

Delta = b2 - 4ac = (-15)^2 - 4*(-16)*151 = 9889

sqrt(Delta) = 99.4435

t1 = [-b + sqrt(Delta)] / 2a = (15 + 99.4435) / (-32) = -3.5764 seconds (not useful result, because there is not a negative time)

t2 = [-b - sqrt(Delta)] / 2a = (15 - 99.4435) / (-32) = 2.6389 seconds

Rounding to the nearest hundredth: t = 2.64 seconds

x

step-by-step explanation:

i would you but i don't under stand the qustion

step-by-step explanation: