The height of a ball thrown directly up with a velocity of 40 feet per second from a initial height of of 5 feet is given by the equation h(t)=-16t^2+40t+5, where t is the time in seconds and h is the ball’s height, measured in feet. When will the ball hit the ground? Round your answer to the nearest tenth.

When it hits the ground , h(t) = 0 , so

-16t^2 + 40t + 5 = 0
16t^2 - 50t - 5 = 0

Now solve this quadratic using the method that you learned.
One of the answers will be negative, you would reject that one.

2.5

To find out when the ball will hit the ground, we need to determine the time at which its height is zero.

In the given equation, the height of the ball is represented by h(t) = -16t^2 + 40t + 5.

Setting h(t) to zero, we get:
-16t^2 + 40t + 5 = 0

This is a quadratic equation. To solve it, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -16, b = 40, and c = 5. Substituting these values into the quadratic formula, we have:
t = (-(40) ± √((40)^2 - 4(-16)(5))) / (2(-16))

Simplifying this expression, we get:
t = (-40 ± √(1600 + 320)) / (-32)
t = (-40 ± √(1920)) / (-32)

Now, we can find the solutions for t by evaluating both values of the expression above:
t1 = (-40 + √(1920)) / (-32)
t2 = (-40 - √(1920)) / (-32)

Calculating these values, we get:
t1 ≈ 0.724 (rounded to three decimal places)
t2 ≈ 2.276 (rounded to three decimal places)

The ball will hit the ground when t = 2.276 seconds (rounded to the nearest tenth).

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