Answer:
F = 20.4 i ^
Explanation:
This exercise can be solved using the ratio of momentum and amount of movement.
I = F t = Dp
Since force and amount of movement are vector quantities, each axis must be worked separately.
X axis
Let's look for speed
cos 45 = vₓ / v
vₓ = v cos 45
vₓ = 8 cos 45
vₓ = 5,657 m / s
We write the moment
Before the crash p₀ = m vₓ
After the shock
= -m vₓ
The variation of the moment Δp = mvₓ - (-mvₓ) = 2 m vₓ
The impulse on the x axis Fₓ t = Δp
Fₓ = 2 m vₓ / t
Fx = 2 0.450 5.657 / 0.250
Fx = 20.4 N
We perform the same calculation on the y axis
sin 45 = vy / v
vy = v sin 45
vy = 8 sin 45
vy = 5,657 m / s
We calculate the initial momentum po = m 
Final moment
= m
Variations moment Δp = m
- m
= 0
Force in the Y-axis
= 0
Therefore the total force is
F = fx i ^ + Fyj ^
F = Fx i ^
F = 20.4 i ^
I believe the answer would be "Different Colors," hope this helps :)
Answer:
The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Explanation:
Using the equations of motion,
When the rocket is fired from the ground,
u = initial velocity = 0 m/s (since it was initially at rest)
a = 10 m/s²
The engine cuts off at y = 0.5 km = 500 m
The velocity at that point = v
v² = u² + 2ay
v² = 0² + 2(10)(500) = 10000
v = 100 m/s
The velocity at this point is the initial velocity for the next phase of the motion
u = 100 m/s
v = final velocity = 0 m/s (at maximum height, velocity = 0)
y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?
g = acceleration due to gravity = - 9.8 m/s²
v² = u² + 2gy
0 = 100² + 2(-9.8)(y)
- 19.6 y = - 10000
y = 510.2 m = 0.510 km
So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Hope this helps!!!