Question

A cannonball is shot horizontally off a high castle wall at 47.4 m/s. What is the magnitude of the cannonball's velocity after 1.23 s? (Ignore direction)

Answer 1

V = 49.2 m/s

Step-by-step explanation:

Given:

Vₓ = 47.4 m/s

t = 1.23 s

___________

V - ?

Vy = Voy + g·t = 0 + 9.8·1.23 ≈ 12,1 m/s

V = √( (Vx)² + (Vy)² ) = √ ( 47,7² + 12,1²) = 49.2 m/s

Answer 2

48.9 m/s

Step-by-step explanation:

I am ignoring air resistance in this:

horizontal velocity stays constant, so we need to find vertical velocity

v_0y = 0 m/s, t = 1.23s, a = 9.81 m/s^2 <- (g)

v_y = v_0y + at = 0 + 9.81 * 1.23 = 12.066 m/s

the magnitude v = sqrt(v_x ^ 2 + v_y ^2) = sqrt(47.4^2 + 12.066^2) = 48.9m/s