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A.1 · 9d · A.1 Kinematics · projectile motion at an angle, components, s = ut + ½at²

Exam question IV: a ball thrown over a wall

Launch at an angle. Resolve into components, find the time, then the height. Then change every number and finish with one of three questions.

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Hook

A student throws a ball at a wall. Will it get over? If the wall were further away, would it still? And if it hits, how high up? One set of equations answers all three.

The question · 7 marks
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A ball thrown over a wall

1.2 m40°22 m s⁻¹h = ?the ball justgoes over the wall36 m(i) time to reach the wall = ?(ii) u_y = ? (iii) h = ?drawn to scale · air resistance negligible
The situation, drawn to scale. The dashed curve is the path; the three unknowns are the time to the wall, the vertical component u_y of the launch velocity, and the wall height h.

A student throws a ball over a vertical wall. The ball leaves the hand 1.2 m above level ground with a speed of 22 m s⁻¹, at 40° above the horizontal. The wall is a horizontal distance of 36 m from the student. Air resistance is negligible.

PartTaskMarks
(a)Define velocity1
(b)(i)Determine the time taken for the ball to reach the wall2
(b)(ii)Calculate the vertical component u of the initial velocity of the ball1
(b)(iii)The ball just goes over the wall. Calculate the height h of the wall3
Check 1 of 7

A ball is thrown at 22 m s⁻¹ at 40° above the horizontal. The initial vertical component of its velocity is…

Check 2 of 7

At the highest point of its flight (no air resistance), the vertical component of the ball’s velocity is…

Check 3 of 7

A ball is thrown from the ground at 20 m s⁻¹, 30° above the horizontal, toward a wall 30 m away. The time to reach the wall is about…

Check 4 of 7

For the same throw (g = 9.81 m s⁻²) the ball’s height above the ground when it reaches the wall is…

Check 5 of 7

The ball “just goes over” a wall. This means that at the wall, the height of the ball…

Check 6 of 7

In part (iii), rounding the time 2.136 s to 2.1 s changes the height by about 0.3 m. The main reason is that…

Check 7 of 7

The vertical displacement of the ball from the launch point, when it reaches the wall, is 7.83 m. The wall’s height above the ground is…

Apply

The student throws at the same angle, 40°, but a little faster. The wall stays 36 m away. Explain, using the equations, what happens to the height of the ball at the wall.

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Remember earlier lessons
From lesson A.1 · 9c

The horizontal component of a projectile’s velocity (no air resistance) is…

From lesson A.1 · 9b

A ball is dropped and another is thrown sideways from the same height. They…

Summary card

Key points

  • • Resolve the launch velocity: u_x = u cosθ, u_y = u sinθ. Fix a sign convention: up is +, so a_y = −g.
  • • Time comes from the horizontal axis: t = d / u_x. Height comes from the vertical axis: y = y₀ + u_y t − ½gt².
  • • Add the release height to get the height above the ground. Keep the unrounded time.
  • • Three finishing questions: will it clear (compare), how far should the wall be (solve y = h, take the later root), how high does it hit (y at the wall).

Formulas

  • ux = u cosθ
  • uy = u sinθ
  • t = d / ux
  • y = y0 + uyt − ½gt²

Key terms

Velocity
:
rate of change of displacement, a vector
Component
:
the part of a vector along one axis

Common errors

  • • Using 22 m s⁻¹ as a component.
  • • Forgetting to add the release height.
  • • Rounding the time too early.
  • • Mixing the sign of g.
Checks solved: 0 of 7
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