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.
Well, Patty may only be 7 years old, but she's aiming high! Let's calculate how high she can throw the ball by plugging in the values into the equation.
Given:
v = 11 ft/s (initial velocity)
s = 0 ft (initial height)
The equation for the height of the ball becomes y = -16t^2 + 11t.
To find out how high Patty can throw the ball, we need to find the maximum height, which occurs at the vertex of the parabolic shape. The formula for finding the x-coordinate of the vertex is given by -b/2a, in this case, -v/2a.
Since a = -16, and v = 11, we can substitute these values into the formula to find the time when the ball reaches its maximum height:
t = -11 / (2 * -16) = 11/32 ≈ 0.344 sec.
Next, we substitute t = 0.344 sec back into the equation y = -16t^2 + 11t to find the maximum height:
y = -16(0.344)^2 + 11(0.344) ≈ -1.816 + 3.784
y ≈ 1.968 ft.
So, Patty can throw the ball approximately 1.968 feet high. Good for her! Although, she might need a boost to reach a regulation-height hoop.
To find out how high Patty can throw the ball, we need to calculate the maximum height of the ball.
The equation for the height of the ball is given as y = -16t^2 + vt + s, where v is the initial velocity, t is time in seconds, and s is the initial height.
Given that Patty can throw the ball with an initial velocity of 11 feet per second, we can substitute v = 11 into the equation.
The equation becomes y = -16t^2 + 11t + s.
Since we need to find the maximum height, we can find the maximum point of the equation by finding the vertex of the parabolic equation. The vertex of a parabola in the form y = ax^2 + bx + c is given by the formula x = -b / 2a.
In this case, a = -16 and b = 11. Plugging these values into the formula, we get:
t = -11 / (2 * -16)
t = -11 / -32
t = 0.34375 seconds.
To find the maximum height, we substitute this value of t back into the equation:
y = -16(0.34375)^2 + 11(0.34375) + s.
Simplifying:
y = -16(0.11767578125) + 3.78125 + s
y = -1.8828125 + 3.78125 + s
y = 1.8984375 + s.
The maximum height of the ball is given by this equation, 1.8984375 + s where s is the initial height.
Without knowing the initial height, s, we cannot determine the exact maximum height Patty can throw the ball. If you provide the value of s, we can calculate the maximum height.
To determine how high Patty can throw a ball, we need to solve the equation y = -16t^2 + vt + s, where y is the height of the ball, v is the initial velocity, t is the time in seconds, and s is the initial height.
In this case, Patty's initial velocity, v, is 11 feet per second. However, we need to consider the initial height, s. If Patty is throwing the ball from the ground, the initial height, s, is 0.
Therefore, the equation becomes:
y = -16t^2 + 11t + 0
To find the maximum height, we need to determine the vertex of the parabola represented by the equation. The x-coordinate of the vertex represents the time at which the ball reaches its maximum height.
The x-coordinate of the vertex can be found using the formula:
t = -b / (2a)
In our equation, a = -16 and b = 11. Substituting the values, we get:
t = -11 / (2(-16))
t = -11 / (-32)
t = 11/32
So, the ball reaches its maximum height 11/32 seconds after being thrown.
To find the maximum height (y-coordinate of the vertex), we substitute this time value back into the equation:
y = -16(11/32)^2 + 11(11/32) + 0
Simplifying the equation gives us the maximum height:
y = -193/16 + 121/32
y = -386/32 + 121/32
y = -265/32
Therefore, Patty can throw the ball to a maximum height of -265/32 feet. However, since height cannot be negative, we can conclude that the maximum height Patty can throw the ball is 8.28125 feet (approximately).
In the form of the equation you are using, v is the initial velocity, so
y = -16t^2 + 11t + s
The maximum height would be obtained at the vertex of this parabola.
the x of the vertex is -11/-32 seconds or .34 seconds
then y = -16(.34)^2 + 11(.34) + s = 1.89 + s
unless she is about 8 feet tall, she is not going to make it.
(s will be the initial height, probably the position of her hand)