Answer:
a) j(t) = -16t^2 +20t +6
b) 1 second
c) less time
Step-by-step explanation:
The general equation for vertical motion is ...
[tex]h(t)=-\dfrac{1}{2}gt^2+v_0t+h_0\\\text{where $v_0$ is the initial upward velocity, and $h_0$ is the initial height}[/tex]
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a) j(t) = -16t^2 +20t +6 . . . . . filling in the given initial values, with g=32 ft/s^2
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b) We want to find the larger of the solutions to ...
j(t) = 10 = -16t^2 +20t +6
Subtracting 6 and dividing by -4, we have ...
-1 = 4t^2 -5t
4t^2 -5t +1 = 0 . . . . add 1, put in standard form
(4t -1)(t -1) = 0 . . . . . factor
t = 1/4 or 1
The ball is in the air 1 second before reaching the hoop.
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c) In generic terms, the equation of motion for a given shooting height at the same upward velocity will be ...
10 = -16t^2 +20t + h0
-16t^2 +20t +(h0 -10) = 0
Using the quadratic formula, we find the larger solution to be ...
t = (-(20) -√(20² -4(-16)(h0 -10)))/(2(-16))
t = (5/8) + √(64h0 -240)/32 = (5 +√(4h0 -15))/8
This value increases as h0 increases, so the ball will be in the air less time when Raymond shoots.
