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# The path of a basketball (maximum height, horizontal distance etc.)?

I've honestly never understood these questions and my teacher sucks at explaining

The path of a basketball shot can be modelled by the equation h=-0.125d^2 + 2.5, where h is the height of the basketball, in metres, and d is the horizontal distance of the ball from the player, in metres.

a) find the maximum height reached by the ball

b) what is the horizontal distance of the ball from the player when it reaches its maximum height?

c) how far from the floor is the ball when the player releases it?

### 2 Answers

Relevance
• Given: h = -0.125d² + 2.5

a) max height when dh/dt = 0

dh/dt = 0 = -0.25d

true when d = 0,

making the maximum height 2.5 m.

(I wonder if you left a term out of your equation? That shot never reaches the basket. I suspect your equation should read either

h = -0.125(d - k)² + 2.5

or h = -0.125d² + k*d + 2.5

where k is some number. IF that's right, then

dh/dt = 0 = -0.25d + k

would let you solve for d --

d = k / 0.25

then plug d into the main equation.)

b) d = 0 m

c) h = 2.5 m

• Put h=-0.125d^2 + 2.5 into full quadrativ form:

h=-0.125d^2 + 0d + 2.5

Then use 1st part of QF d= -b/2a, in this case d= 0 ((which doesn't make any sense))

Max height is then h = 2.5m ((which doesn't make any sense again))

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