Respuesta :
Answer:
x = 32356.44 m
|V | = 567,02 m/s
θ = 42 ( second quadrant )
Step-by-step explanation:
With V₀ = 570 m/s and θ = 40° ( angle above horizontal)
we get
Vox = Vo* cos θ Vox = 570 * 0,7660 Vox = 432.62 m/s
Voy = Vo* sin θ Voy = 570 * 0.6428 Voy = 366,40 m/s
Just at the moment of ground impact y = 0
y = y₀ + Voy * t - 1/2 g*t²
By subtitution
0 = 6 + 366,40*t - 4.9*t²
In this second degree equation we solve for x
4.9*t² - 366.40*t - 6 = 0
t₁,₂ = [ 366.40 ± √ (366.40)² + 117.6] / 9.8
t₁ = -0.16/9.8 we dismiss this negative root
t₂ = 74.79 s which is the total time of the movement
Then x the horizontal displacement is:
x = Vox * t x = 432.62 * 74.79 x = 32356.44 m
To find the speed of the projectile hitting the ground
Vy = Voy - g*t Vy = 366.40 - 9.8 * 74.79 Vy = 366,54 m/s
Then V = √ (Vx)² + (Vy)² V = √ (432.62)² + (366.54)²
|V | = 567,02 m/s
and the angle is
θ = arctan ( Vy/Vx) θ = arctan - 366.54/432,62 θ = arctan -0.85
θ = 42 ( angle between the projectile and the horizontal axis (second quadrant)
