Asked by uriel
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.
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.
Answers
Answered by
Reiny
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)
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)
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