Martin throws a ball straight up in the air. the equation h(t) = -16t^2 + 40t + 5 gives the height of the ball, in feet, t seconds after martin releases it.
how many seconds before the ball martin threw hits the ground?
show your work.
(pls answer asap!!)

2.62 seconds

Step-by-step explanation:

Let

t ----> the time in seconds

h(t) ----> he height of the ball, in feet

we have

we know that

When the ball hits the ground, the height is equal to zero

so

The formula to solve a quadratic equation of the form

is equal to

in this problem we have

so

substitute in the formula

therefore

The solution  is t=2.62 seconds

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step-by-step explanation:

add -60 to 120 which will result in : 2x+3y=180

then subtract 2x: 3y=-2x+180

lastly, divide by 2: y= -2/3x+60

hope this is the answer you're looking for: )