A flaming arrow is shot into the air to mark the beginning of the a festival. Its height, h, in meters after t seconds can be modelled by the function h(t)=-4.9t^2+24.5t+2

How fast is the arrow traveling when it hits the ground? Explain how you arrived at your answer.

(prior to that i found the time it takes for the arrow to reach the ground level a 5.0803s)

The way to solve this is explained in your subject title. Compute the derivative of the function, which is the speed, and calculate its value at t = 5.0803 s.

The derivative is
V(t) = dh/dt = -9.8t + 24.5

Plug in the value of t that you got (which was correct) and compute V

To find the speed of the arrow when it hits the ground, we need to determine the derivative of the function h(t) with respect to time (t). The derivative will give us the rate at which the height is changing with respect to time.

First, let’s write the function h(t) in standard form:
h(t) = -4.9t^2 + 24.5t + 2

Now, to find the derivative, we need to differentiate each term in the equation. The derivative of -4.9t^2 is -9.8t, and the derivative of 24.5t is 24.5. Since the derivative of a constant (2 in this case) is zero, it does not contribute to the derivative. So, the derivative of h(t) is:

h'(t) = -9.8t + 24.5

Now, we have the derivative of the height function with respect to time. In order to find the speed of the arrow when it hits the ground, we need to substitute the time it takes for the arrow to reach the ground into the derivative.

Given that the time it takes for the arrow to reach the ground is 5.0803 seconds (as you mentioned), we can substitute this value into the derivative:

h'(5.0803) = -9.8(5.0803) + 24.5

Calculating this expression will give us the speed of the arrow when it hits the ground.

To find the speed at which the arrow is traveling when it hits the ground, we need to calculate its velocity at that moment. Velocity is the derivative of the height function with respect to time.

Given the height function h(t) = -4.9t^2 + 24.5t + 2, we can find the derivative of this function to get the velocity function.

The derivative of h(t) with respect to t can be found by applying the power rule of differentiation. The power rule states that for a function of the form f(t) = a*t^n, the derivative is given by f'(t) = a*n*t^(n-1). Applying this rule to our height function:

h'(t) = -4.9 * 2 * t^(2-1) + 24.5 * 1 * t^(1-1) + 0
h'(t) = -9.8t + 24.5

So, the velocity function is v(t) = -9.8t + 24.5.

Now, we need to find the time t when the arrow hits the ground. You have already found this value to be 5.0803 seconds.

To find the speed, we substitute this time value into the velocity function: v(5.0803) = -9.8 * 5.0803 + 24.5.

Evaluating this expression, we find that the speed of the arrow when it hits the ground is approximately 21.725 meters per second.

Therefore, the arrow is traveling at a speed of approximately 21.725 m/s when it hits the ground.