Projectile Motion
1 answer:
Answer:
a) F = (m / t₀) 95.33
b) θ = 70.5º
Explanation:
This is a projectile launch, as they indicate the horizontal distance this is the range of the body, let's use the expression for the range of the projectiles
R = v₀² sin 2θ / g
v₀² sin 2θ = R g
Where the range is 550.46 m
They also indicate the time that the air must remain, so this time is twice the time until reaching the maximum height.
v_{y} = - g t
At the maximum height v_{y} = 0 and the initial speed on the axis and we can find it with trigonometry
sin θ = v_{oy} / v_{o}
v_{oy} = v_{o} sin θ
v_{o} sin θ = g t
Let's write the two equations
v_{oy}² sin 2θ = g R
v_{o} sin θ = g t
We solve our accusation system
(G t / sin θ) 2 sin 2θ = g R
g t² sin 2θ = R sin θ
Let's use the trigonometric relationship
sin 2θ = 2 sin θ cos θ
We substitute
g t² (2 sin θ cos θ) = R sin θ
Cos tea = R / 2 g t²
θ = cos⁻¹ (R / 2g t²)
Let's calculate
θ = cos⁻¹ (550.46 / (2 9.8 9.17² ))
θ = 70.5º
a) Force can be Newton's second law
On the x axis the speed is constant so the force on the axis is zero
In the y axis the acceleration we have is the acceleration of gravity, so the force that acts throughout the journey is the weight of the body.
To place the body in the air from the rest we can use the equation of the impulse
F t = Δp = m v - m v₀
As kick from rest v₀ = 0
Let's find the speed of the body
= g t
v_{o} = g t / sin 70.5
v_{o} = 9.8 9.17 / sin 70.5
v_{o} = 95.33 m / s
To encocorate the force we must suppose a firing time, which in general is very short, suppose that this time is to
F = m v_{o} / t₀
F = (m / t₀) 95.33
This is the outside that should be applied, as an example suppose a body of mass 1 kg⁵ ( m = 1 kg) and a trip time to = 0.1 s
F = (1 / 0.1) 95.33
F = 953.3 N
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