Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s
Answer:
L/2
Explanation:
Neglect any air or other resistant, for the ball can wrap its string around the bar, it must rotate a full circle around the bar. This means the ball should be able to swing to the top position where it's directly above the bar. By the law of energy conservation, this happens when the ball is at the same level as where it's previously released vertically. It means the swinging radius around the bar must be at least half of the string length.
So the distance d between the bar and the pivot should be at least L/2
A. 10 rations = 1 deca-ration.
b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.
c. 10⁻⁵ phones = 1 micro-phones.
d. 10⁻⁹ goats = 1 nano-goats.
e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.
a) 2.75 s
The vertical position of the ball at time t is given by the equation
where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:
This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:
where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:
And the negative sign means the direction is downward.
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