# Tan line and velocity problems

The parabola y = (x^2) + 3 has two tangents which pass through the point (0, -2). One is tangent to the to the parabola at (A, A^2 + 3) and the other at (-A, A^2 + 3). Find (the positive number) ?

If a ball is thrown vertically upward from the roof of 64ft foot building with a velocity of 96 ft/sec, its height after t seconds is s(t) = 64 + 96t - 16t^2. I found it's max height to be 208 ft. What is the velocity of the ball when it hits the ground (height0)?

Use the derivative to find this equation of the tangent line to the curve y = 4x + 3 * square root of x at the point (4, 22.000000). The equation of this tangent line can be written in the form y = mx + b?

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#### Solution Preview

...ing 1 and 2, we get:
{A^2+5}/A = 2A
A = 5^0.5 = 2.236
(2)
The most effective equation here is:
v^2 = u^2 + 2gh
First we have to calculate the maximum height the ball reaches. Since its going against the gravity v = 0, u = 96 ft/s and g = -32 ft/s^2
0 = 96^2 - 2*(-32)*h
h = 144 ft. Now ball will stop at ...