A baseball is thrown in a parabolic arc. It's position above the ground at a given point in time can be represented by the quadratic function LaTeX: p\left(t\right)=\frac{1}{2}gt^2+v_0 t+p_0p ( t ) = 1 2 g t 2 + v 0 t + p 0, where LaTeX: t\ge0t ≥ 0, LaTeX: gg is -32 ft/sec/sec, LaTeX: v_0v 0 is initial velocity, and LaTeX: p_0p 0 is its initial position above the ground. If the ball was thrown straight up at 16 ft/sec when it was 5 ft above the ground, how high did it go?

The baseball reached 9 feet high above the ground

Step-by-step explanation:

The given quadratic function representing the position of the baseball above ground is p(t) = 1/2·g·t² + v₀·t + p₀

Where;

t ≥ 0

g = -32 ft./sec²

v₀ = The initial velocity

p₀ = The initial position

Given that when the ball is thrown, we have;

The initial, straight up, velocity, v₀ = 16 ft./sec

The initial position, p₀ = 5 ft.

Substituting the above values in the quadratic function representing the position of the baseball above ground, we have;

p(t) = 1/2·(-32)·t² + 16·t + 5 = 16·t - 16·t² + 5

At the maximum point, the rate of change of the height with time = 0, therefore;

dp(t)/dt = 0 = d(16·t - 16·t² + 5)/dt = 16 - 32·t

16 - 32·t = 0

16 = 32·t

t = 16/32 = 0.5 seconds

Therefore, the time takes to reach the maximum height = 0.5 seconds

The height (maximum) reached in 0.5 seconds is given as follows;

h(t) = 16·t - 16·t² + 5, from which we have;

h(0.5) = 16 × 0.5 - 16 × (0.5)² + 5 = 9

Therefore, the height baseball reached = 9 ft. above ground

x = 25

step-by-step explanation:

a^2+b^2=c^2

7^2+24^2=c^2

49+576=625

sqrt625=25

x=25