# an object is thrown upward from a height of 70 feet. t second after it is released, the object is h=-16t^2+40t+70 feet above the ground. how many seconds does it take until the object hits the ground?

## Tr = -Vo/g = -40/-32 = 1.25 s. = Rise

time or time to reach max ht.

h = -16*1.25^2 + 40*1.25 + 70 = 95 Ft.

h = 0.5g*t^2 = 95
16t^2 = 95
t^2 = 5.94
Tf = 2.44 s. = Fall time.

Tr+Tf = 1.25 + 2.44 = 3.69 s. = Time to
reach gnd.

## To find the time it takes for the object to hit the ground, we need to determine when the height h equals zero. In this case, h represents the height of the object above the ground.

The equation h = -16t^2 + 40t + 70 represents the height of the object as a function of time (t).

Setting h equal to zero, we get:

0 = -16t^2 + 40t + 70

Now, we need to solve this quadratic equation for t. We can do this by factoring, completing the square, or using the quadratic formula. Let's solve it using the quadratic formula:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In our case, a = -16, b = 40, and c = 70. Plugging these values into the quadratic formula, we have:

t = (-40 ± sqrt(40^2 - 4(-16)(70))) / (2(-16))

Simplifying further:

t = (-40 ± sqrt(1600 + 4480)) / (-32)

t = (-40 ± sqrt(6080)) / (-32)

Now, we evaluate the square root:

t = (-40 ± sqrt(6080)) / (-32)

t = (-40 ± 77.96) / (-32)

Now, we have two possible solutions for t:

t1 = (-40 + 77.96) / (-32) ≈ 0.87 seconds (rounded to two decimal places)

t2 = (-40 - 77.96) / (-32) ≈ -2.71 seconds (rounded to two decimal places)

Since time cannot be negative in this context, we discard the negative value. Therefore, the object will hit the ground approximately 0.87 seconds after it was released.