← Horizontal launches: the two problems in numbers

Velocity at a point in flight

Two perpendicular displacements: 7 m south and 24 m east. Enter the resultant's magnitude in m.

Two perpendicular displacements: 20 m east and 11 m north. Enter the resultant's angle north of east, to the nearest degree. Answers within 1° are accepted.

Partway through its flight, a ball launched horizontally from a rooftop moves sideways at 12 m/s and downward at 5 m/s. The ball has one velocity, not two. What are its speed and direction at that moment?

The sketch from the last lesson showed the two component arrows at every point: an unchanging horizontal arrow, and a vertical arrow that free fall grows. Speed and direction come from recombining the two arrows: magnitude by the Pythagorean theorem, direction by the tangent.

The two components are perpendicular, so the speed is the hypotenuse of their right triangle:

v = √(12² + 5²) = √169 = 13 m/s

The direction comes from the tangent of the angle below the horizontal:

tan θ = 5 ÷ 12, so θ = tan⁻¹(5 ÷ 12) ≈ 23°

v_x = 12 m/sv_y = 5 m/sv = 13 m/s23°
At this moment the components are 12 m/s sideways and 5 m/s down. Together they are one velocity: 13 m/s at 23° below the horizontal.
The ball moves at 13 m/s, angled 23° below the horizontal. The angle is measured down from the horizontal because the vertical component points down.

A stone thrown horizontally from a bridge moves, at one moment, 8 m/s horizontally and 6 m/s downward. Calculate its speed at that moment.

Write down the values in the question:

v_x = 8 m/s

v_y = 6 m/s downward

v = ?

Write down the equation:

v = √(v_x² + v_y²)

Substitute in the values:

v = √(8² + 6²) = √100

Simplify:

v = 10.0 m/s

The same stone: v_x = 8 m/s, v_y = 6 m/s downward. Calculate the angle of its velocity below the horizontal.

Write down the values in the question:

v_x = 8 m/s

v_y = 6 m/s downward

θ = ?

Write down the equation:

tan θ = v_y ÷ v_x

Substitute in the values:

tan θ = 6 ÷ 8

Simplify:

θ = tan⁻¹(6 ÷ 8) = 36.9° below the horizontal

Late in its flight off a cliff, a ball moves 20 m/s horizontally and 21 m/s downward. Enter its speed at that moment in m/s, to one decimal place.

The same moment: v_x = 20 m/s, v_y = 21 m/s downward. Enter the angle of the ball's velocity below the horizontal, in degrees to one decimal place.

Early in its flight off a rooftop, a baseball moves 24 m/s horizontally and 7 m/s downward. Enter its speed at that moment in m/s, to one decimal place.

Your turn

At one moment of its flight off a pier, a fishing float moves 9 m/s horizontally and 12 m/s downward. Enter its speed at that moment in m/s, to one decimal place.

Shortly after leaving a shelf, a marble moves 5.0 m/s horizontally and 2.0 m/s downward. Enter the angle of its velocity below the horizontal, in degrees to one decimal place.

A ball leaves a cliff top horizontally at 10 m/s. It is still in the air 2.0 s later. Enter its speed at that moment in m/s, to one decimal place.