Question

an object that is launched from the top of a tall tower at and angle has initial velocity has components (vx = 30 m/s, vy = 20 m/s). ignore the effects of air resistance
what are the horizontal and vertical components of the objects velocity 3 seconds after launch?

Explanation:

Step1: Analyze horizontal velocity

In projectile motion (ignoring air - resistance), the horizontal velocity \(v_x\) remains constant. Given \(v_{x0}=30m/s\), so \(v_x = 30m/s\) at \(t = 3s\).

Step2: Analyze vertical velocity

Use the formula \(v_y=v_{y0}-gt\) (taking upward as positive, \(g = 10m/s^{2}\), \(v_{y0}=20m/s\), \(t = 3s\)).
Substitute the values: \(v_y=20-10\times3\).
Calculate: \(v_y=20 - 30=- 10m/s\).

Answer:

C. Horizontal Velocity (\(Vx\)) \(30m/s\), Vertical Velocity (\(Vy\)) \(-10m/s\)