Answer:
The airplane should release the parcel m before reaching the island
Explanation:
The height of the plane is , and its speed is v=150 m/s
When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is
[1]
And the distance X is
x = V.t [2]
Being t the time elapsed since the release of the parcel
If we isolate t from the equation [1] and replace it in equation [2] we get
Using the given values:
x = m
Answer:
v = 1176.23 m/s
y = 741192.997 m = 741.19 km
Explanation:
Given
M₀ = 9 Kg (Initial mass)
me = 0.225 Kg/s (Rate of fuel consumption)
ve = 1980 m/s (Exhaust velocity relative to rocket, leaving at atmospheric pressure)
v = ? if t = 20 s
y = ?
We use the equation
v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt where t ∈ (0, t)
⇒ v = - ve*Ln ((M₀ - me*t)/M₀) - g*t
then we have
v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)
v = 1176.23 m/s
then we apply the formula
y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt
⇒ y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt
⇒ y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀ - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)
For t = 20 s we have
y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)
⇒ y = 741192.997 m = 741.19 km
The graphs are shown in the pics.
Answer:
Explanation:
As we know that net displacement is
altitude is given as
so its horizontal displacement is given as
now we have
now in x direction we have
in y direction we have
from above equations we have
by solving above equation