Question

A projectile is fired with an initial muzzle speed 570 m/s at an angle 40∘ from a position 6 meters above the ground level. Find the horizontal displacement from the firing position to the point of impact. At what speed does the projectile hit the ground?

Answer 1

x = 32356.44 m

|V | = 567,02 m/s

θ = 42 ( second quadrant )

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)