Substituting eq. (8) back into eq. (5). gives an expression for the minimum initial velocity consistent with the ball reaching the hoop. The result is ¹0 = 8√ H+ √H² + x²) where = H-h (9)
Substituting eq. (8) back into eq. (5). gives an expression for the minimum initial velocity consistent with the ball reaching the hoop. The result is ¹0 = 8√ H+ √H² + x²) where = H-h (9)
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Can someone explain this why the answer became like this? And show me the step-by-step solution please
![Here g is acceleration due to gravity at the earth's surface. t, and t₂ are time interval when the ball
is released from the player to the maximum height (Y) and the time interval from the maximum height
to the hoop respectively.
To find the minimum initial velocity of the ball, the relationship between release angle and initial
velocity for a given range must be considered. From eq. (3), we can be rewritten as
H = (₁ +₂126₁-(₁+₂)].
2
Substituting eq. (1) and eq. (2) in eq. (4) we got
X
X
21,
V cos
H-B..
2 v cos
We can rearrange eq. (4) as
gX² (tan²0+1)
2(X tano - K)
The trigonometric identities sec² 0 =
(5)
To solve the problem, we need to differentiate vo with respect to and set the derivative equal to
zero. Then, we got
dv
dvo
d0
where H = H-h.
X tan
X
=(Xtano-H)tano-
2
tan²0+1 enable us to write eq. (6) more simply as
8X²
2v cos²0
cos²0
10
(4)
sec² 0 = 0.
-=X tan²0-21 tane - X=0.
(6)
- B± √B²-4AC
From quadratic equation Ax² + Bx+C =0, x=-
where AB and Care
2 A
constant. We choose appropriate solution as positive solution, then we get optimal angle as
H+√√H² + X²
0= tan
(7)
(8)
Substituting eq. (8) back into eq. (5). gives an expression for the minimum initial velocity
consistent with the ball reaching the hoop. The result is
-√H²+
(9)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7b9b716-1b90-48a7-9ec9-5173a2e757ce%2F2647abe8-50fc-49e1-97c3-ef1425cee3f8%2F1wc7ygfc_processed.png&w=3840&q=75)
Transcribed Image Text:Here g is acceleration due to gravity at the earth's surface. t, and t₂ are time interval when the ball
is released from the player to the maximum height (Y) and the time interval from the maximum height
to the hoop respectively.
To find the minimum initial velocity of the ball, the relationship between release angle and initial
velocity for a given range must be considered. From eq. (3), we can be rewritten as
H = (₁ +₂126₁-(₁+₂)].
2
Substituting eq. (1) and eq. (2) in eq. (4) we got
X
X
21,
V cos
H-B..
2 v cos
We can rearrange eq. (4) as
gX² (tan²0+1)
2(X tano - K)
The trigonometric identities sec² 0 =
(5)
To solve the problem, we need to differentiate vo with respect to and set the derivative equal to
zero. Then, we got
dv
dvo
d0
where H = H-h.
X tan
X
=(Xtano-H)tano-
2
tan²0+1 enable us to write eq. (6) more simply as
8X²
2v cos²0
cos²0
10
(4)
sec² 0 = 0.
-=X tan²0-21 tane - X=0.
(6)
- B± √B²-4AC
From quadratic equation Ax² + Bx+C =0, x=-
where AB and Care
2 A
constant. We choose appropriate solution as positive solution, then we get optimal angle as
H+√√H² + X²
0= tan
(7)
(8)
Substituting eq. (8) back into eq. (5). gives an expression for the minimum initial velocity
consistent with the ball reaching the hoop. The result is
-√H²+
(9)
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