Patty is only 7 years old, but she insists she can play basketball with a regulation-height hoop. She can throw the ball with an initial velocity of 11 feet per second.
The equation for the height of the ball is y = −16t2 + vt + s
where v is the initial velocity, t is in seconds, and s is the initial height.
How high can she throw a ball? Answer in units of ft.
23
To find out how high Patty can throw a ball, we need to substitute the given values into the equation for the height.
Given:
Initial velocity (v) = 11 feet per second
Initial height (s) = 0 feet (assuming Patty is throwing the ball from the ground)
The equation for the height is:
y = -16t^2 + vt + s
Substituting the given values:
y = -16t^2 + 11t + 0
Since we want to find the maximum height the ball reaches, we need to find the vertex of the parabolic equation. The vertex of a quadratic equation in the form of y = ax^2 + bx + c is given by the formula:
x = -b / 2a
In our case:
a = -16
b = 11
Substituting the values:
t = -11 / 2*(-16)
t = 11 / 32
Now we have the time (t) at which the ball reaches its maximum height. We can substitute this value back into the original equation to find the height (y).
Substituting t = 11 / 32:
y = -16(11 / 32)^2 + 11(11 / 32) + 0
Evaluating the expression, we can find the maximum height Patty can throw the ball.