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:
Relative age-dating involves comparing a rock layer or rock structure with other near-by layers or structures. Using the principles of superposition and cross-cutting relationships, and structures such as unconformities, one can determine the order of geological events.
Blood cell : Eukaryotic cell
and
Bacteria : Prokaryotic cell.
Explanation:
Doing homework is risky behaviour broo
Answer: Speed = 4 m/s
Explanation:
The parameters given are
Mass M = 60 kg
Height h = 0.8 m
Acceleration due to gravity g= 10 m/s2
Before the man jumps, he will be experiencing potential energy at the top of the table.
P.E = mgh
Substitute all the parameters into the formula
P.E = 60 × 9.8 × 0.8
P.E = 470.4 J
As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.
When hitting the ground,
K.E = P.E
Where K.E = 1/2mv^2
Substitute m and 470.4 into the formula
470.4 = 1/2 × 60 × V^2
V^2 = 470.4/30
V^2 = 15.68
V = square root (15.68)
V = 3.959 m/s
Therefore, the speed of the man when hitting the ground is approximately 4 m/s