Answer:
58.5 m
Explanation:
First of all, we need to find the total time the ball takes to reach the water. This can be done by looking at the vertical motion only.
The initial vertical velocity of the ball is
where
u = 21.5 m/s is the initial speed
is the angle
Substituting,
The vertical position of the ball at time t is given by
where
h = 13.5 m is the initial heigth
is the acceleration of gravity (negative sign because it points downward)
The ball reaches the water when y = 0, so
Which gives two solutions: t = 3.27 s and t = -0.84 s. We discard the negative solution since it is meaningless.
The horizontal velocity of the ball is
And since the motion along the horizontal direction is a uniform motion, we can find the horizontal distance travelled by the ball as follows:
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50(6)=300.
The ball traveled 300 meters in 50 seconds.
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Answer:
d₁ = 0.29 in
d₂ = 0.505 in
Explanation:
Given:
T = 1500 lbf in
L = 10 in
x = 0.5 L = 5 in
First case: T = T₁ + T₂
T₂ = T - T₁ = 1500 - 750 = 750 lbf in
If the shafts are in series:
θ = θ₁ + θ₂
θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)
Second case: If d₁ ≠ d₂
θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)
t₁ = t₂
(eq. 2)
T₁ + T₂ = 1500 (eq. 3)
θ₁ first case = θ₁ second case
Replacing:
The same way to θ₂:
From equation 2, we have:
d₁ = 0.587 * d₂
From equation 3, we have:
d₂ = 0.505 in
d₁ = 0.29 in
