What is the velocity of a javelin that travels 45 meters in 3 seconds?

Answer 1

1. For the vertical direction, it's a constant acceleration g. For the javelin to fall back to the ground,

S_y = V_y t - 1/2gt**2 = 0, so Vy = 1/2 gt = 15m/s

1. For the horizontal direction, it's a constant velocity. so

S_x = Vxt, so V_x = 45/3 15m/s.

1. So the velocity of the javalin can be obtained by sqrt(V_x ² + V_y² ) = 21.21 m/s