A ball is thrown vertically upward from the ground with an initial velocity of 111 ft/sec. Use the quadratic function h(t) = −16t2 + 111t + 0 to find how long it will take for the ball to reach its maximum height (in seconds), and then find the maximum height (in feet). (Round your answers to the nearest tenth.)

t=3.5 seconds

Step-by-step explanation:

Given

h(t) = −16t^2 + 111t + 0

h'(t)= -32t + 111

Maximum height occurs when h(t) = 0 and the ball begins to fall

h(t)= -32t + 111=0

-32t + 111=0

-32t=-111

Divide both sides by -32

t=3.46872

Approximately, t=3.5 seconds

Recall,

Maximum height occurs when h(t) = 0

h(t)= -32t + 111=0

= -32(3.46872)+111

= -110.99904+111

= 0.00096 ft

because its not ur topic and its not for is for homework not that.

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1