A science project studying catapults sent a projectile into the air with an initial velocity of 24.5 m/s. The formula for distance (s) in meters with respect to time in seconds is

s = -4.9t2 + 24.5t.

a)Find the time that this projectile would appear to have the maximum distance above ground.

B)Find the slope of the tangent at that point using the limit as h approaches 0 of f(x+h) - f(x) over h

The maximum height is the VERTEX. You have the equation, and just need to use grade 10 math : ) for the vertex.

actually I think the 24.5 is the initial velocity component up and 4.9 is (1/2)g downward

If
g = -9.8 m/s^2
then
v = Vi - g t where Vi is initial speed up
and height your s, is height up above starting point
s = Vi t - (1/2) g t^2
s = 24.5 t - 4.9 t^2

Now I could tell you easily when it is at the top. It is when v = 0
so 9.8 t =24.5
but yu are supposed to find the vertex of the parabola I think
4.9 t^2 -24.5 t = -s
t^2 - 5 t = (1/4.9)- s
t^2 - 5 t + 2.5^2 = -(1/4.9)s + 6.25
(t-2.5)^2 = (1/4.9)s + 30.625/4.9 = -(1/4.9)(s-30.625)
so
top at t = 2.5 seconds and s = 30.625 meters
now you know that ds/dt = 0 = v at the top but anyway
you know
s(t+h) = -4.9(t+h)^2 + 24.5 (t+h) = -4.9(t^2+2th+h^2) + 24.5 t + 24.5 h
= -4.9 t^2 - 9.8 t h -4.9 h^2+ 24.5 t + 24.5 h
s(t) = -4.9 t^2 +24.5 t

s(t+h) - s(t) = -9.8 th -4.9h^2 +24.5 h
divide by h to get ds/dt (which I call v :)
ds/dt = -9.8 t -4.9h +24.5 = 24.5 - 9.8 t when h-->0
so I say the speed up = 24.5 - 9.8 t
now when t = 2.5, the top
ds/dt = 24.5 - 9.8(2.5) = 24.5 - 24.5 = 0 Whew !

a) To find the time when the projectile appears to have the maximum distance above the ground, we need to find the vertex of the parabolic equation. The vertex can be found using the formula:

t = -b / (2a)

where a = -4.9 and b = 24.5.

Substituting the values, we get:

t = -24.5 / (2*(-4.9))
t = -24.5 / (-9.8)
t = 2.5 seconds

Therefore, the projectile would appear to have the maximum distance above the ground at t = 2.5 seconds.

b) To find the slope of the tangent at that point, we need to find f(x+h) and f(x), then apply the limit as h approaches 0 to calculate the derivative.

Given that f(x) = -4.9x^2 + 24.5x, let's find f(x+h):

f(x+h) = -4.9(x+h)^2 + 24.5(x+h)
= -4.9(x^2 + 2xh + h^2) + 24.5x + 24.5h
= -4.9x^2 - 9.8xh - 4.9h^2 + 24.5x + 24.5h

Now, we can find the slope of the tangent:

slope = limit as h approaches 0 of (f(x+h) - f(x)) / h

= limit as h approaches 0 of ((-4.9x^2 - 9.8xh - 4.9h^2 + 24.5x + 24.5h) - (-4.9x^2 + 24.5x)) / h

= limit as h approaches 0 of (-9.8xh - 4.9h^2 + 24.5h) / h

= limit as h approaches 0 of (-9.8x - 4.9h + 24.5)

Now, substitute h = 0:

slope = -9.8x + 24.5

Therefore, the slope of the tangent at that point, using the limit as h approaches 0, is -9.8x + 24.5.

a) To find the time when the projectile appears to have maximum distance above the ground, we need to find the vertex of the quadratic equation.

The formula for a quadratic equation in the form of s = at^2 + bt + c can be written as t = -b/2a.

In our equation, s = -4.9t^2 + 24.5t.

Here, a = -4.9 and b = 24.5.

Plugging these values into the formula, we get:
t = -24.5 / (2 * -4.9)
t = 24.5 / 9.8
t ≈ 2.5 seconds

Therefore, the projectile would appear to have the maximum distance above the ground at approximately 2.5 seconds.

b) To find the slope of the tangent at the maximum point, we need to find the derivative of the equation and calculate its value at that point.

The derivative of the equation s = -4.9t^2 + 24.5t is given by ds/dt = -9.8t + 24.5.

Now we substitute the value of t = 2.5 into the derivative:
ds/dt = -9.8 * 2.5 + 24.5
ds/dt = -24.5 + 24.5
ds/dt = 0

The slope of the tangent at the maximum point is 0.