A projectile launched into the air can be modelled by the function h(t) =-4.9t^(2) + 4.1t + 8.2 where h(t) represents the height of the projectile above the ground, in meters, t seconds after it was launched. What time is the projectile at a height of 3 meters?

To find the time when the projectile is at a height of 3 meters, we need to set the given function equal to 3 and solve for t.

So, we have the equation:

-4.9t^2 + 4.1t + 8.2 = 3

Let's solve this equation step by step:

Step 1: Subtract 3 from both sides of the equation:

-4.9t^2 + 4.1t + 8.2 - 3 = 0

This simplifies to:

-4.9t^2 + 4.1t + 5.2 = 0

Step 2: Multiply the entire equation by -10 to eliminate the decimal coefficients:

49t^2 - 41t - 52 = 0

Step 3: We can factor the equation or use the quadratic formula to solve this equation. Let's use the quadratic formula:

The quadratic formula is given by:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 49, b = -41, and c = -52.

t = (-(-41) ± √((-41)^2 - 4(49)(-52))) / (2(49))
t = (41 ± √(1681 + 10192)) / 98
t = (41 ± √(11873)) / 98

Step 4: Calculate the square root:

t = (41 ± 108.92) / 98

Now we have two possible values for t:

t₁ = (41 + 108.92) / 98 ≈ 1.51 seconds
t₂ = (41 - 108.92) / 98 ≈ -0.73 seconds

Since time cannot be negative in this context, we discard the negative value. Therefore, the projectile is at a height of 3 meters after approximately 1.51 seconds.

To find the time when the projectile is at a height of 3 meters, we can set up the equation:

h(t) = 3

Substituting the given function for h(t):

-4.9t^2 + 4.1t + 8.2 = 3

Rearranging the equation:

-4.9t^2 + 4.1t + 8.2 - 3 = 0

Now, we have a quadratic equation of the form:

at^2 + bt + c = 0

where:

a = -4.9
b = 4.1
c = 8.2 - 3 = 5.2

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values:

t = (-(4.1) ± √((4.1)^2 - 4(-4.9)(5.2))) / (2(-4.9))

Simplifying the expression further:

t = (-4.1 ± √(16.81 + 101.12)) / (-9.8)

t = (-4.1 ± √117.93) / (-9.8)

Now, we have two possible values for time, one using the "+" symbol and the other using the "-" symbol:

t = (-4.1 + √117.93) / (-9.8)
t = (-4.1 - √117.93) / (-9.8)

Using a calculator, we can find the approximate values for time:

t ≈ 1.1809 seconds or t ≈ 0.5217 seconds

Therefore, the projectile is at a height of 3 meters approximately 1.1809 seconds and 0.5217 seconds after it was launched.

plug in 3 for h , then solve for t using the quadratic formula