algebra height and velocity

Question

A football player kicks a ball with an initial velocity of 64 feet per second. The initial height of the ball is 4 feet. The function below models the height (in feet) of the ball, when t is the time in seconds.  What is the maximum height the ball reaches? Round to the nearest foot.

Answer 1

So the equation given is h(t) = -16t^2 + 64t + 4. This is a quadratic equation where f(x) = ax^2 + bx + c

It will form a parabola. The x-axis is time and the y-axis is height.

The maximum height is the vertex of the parabola. To find this coordinate, we will need to find the x-coordinate first (which is the "t" in this case).

x-coordinate of vertex = (-b)/(2*a)

Now from the equation, we can see that a = -16 and b = 64. So substituting those values in,

(-64)/ (2 * (-16)) = 64/32 = 2.

So t = 2

Now to find the height, we simply substitute t = 2 into h(t).

h(2) = -16*(2)^2 + 64(2) + 4

= 68

So the maximum height is 68 feet.

Answer 2

h(t) = -16t^2 + 64t + 4

take the derivative and set equal to zero

h'(t) = -32t +64 = 0

t = 64/32 = 2 seconds to reach max height

plug t=2 back into he original equation

h(2) = -16(2)^2 + 64(2) + 4 = -64 + 128 + 4 = 64 +4 = 68 feet high = maximum height

the function h(t) = -16t^2 + 64t + 4

has 3 terms. 1st is the effect of gravity, 2nd is for velocity, 3rd is initial height

A few complications ignored in most solutions to this problem are:

The use of -16t^2 uses 32 feet per second per second which is actually an approximation for the effect of gravity at sea level. A closer approximation is 32.2 ft/sec^2 or 32.174 ft/sec^2. The usual metric equation uses -4.9t^2 which is 9.8 m/sec^2 for the effect of gravity. If you try to convert that to feet per sec per sec, you'd get closer to 32.152 ft/sec^2, but 9.8 is also an approximation. the actual value is coser to 9.80665.

That's at sea level. Altitude reduces the effect of gravity. You weight less at higher altitudes. At 30,000 feet there's a 0.29% reduction in gravitational acceleration, but offset slightly by less air density which increases the effect by 0.08%. Also lattitdue effects gravity. At the poles gravity increases. At the equator it decreases. It's about 0.3% less at the equator. Geography & topology also effect gravity. Knowing the specific city will change the solution too.

Get high enough up, like Engineer/Astronaut Howard (in the Big Bang TV show) and you're seemingly weightless in outer space, but not really, as the apparent "weightlessness" is due to freefall, not the higher altitude where 90% of the gravitational force is still in effect.