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IB MYP 4-5 Physics- Projectile motion- Study Notes

IB MYP 4-5 Physics- Projectile motion- Study Notes- New Syllabus

IB MYP 4-5 Physics- Projectile motion- Study Notes

Key Concepts

  • Projectile Motion

IB MYP 4-5 Physics Study Notes – All topics

Projectile Motion

Projectile Motion

Projectile motion refers to the motion of an object that is thrown or projected into the air and is influenced only by the force of gravity (assuming negligible air resistance). The path followed is a curved parabola.

 

General Features

  • The motion is two-dimensional: horizontal and vertical components.
  • The horizontal component has constant velocity (\( a_x = 0 \)).
  • The vertical component has uniformly accelerated motion (\( a_y = -g \)).
  • The trajectory is parabolic.

Initial Velocity Components

Let the projectile be launched with speed \( u \) at an angle \( \theta \) from the horizontal:

  • Horizontal velocity: \( u_x = u\cos\theta \)
  • Vertical velocity: \( u_y = u\sin\theta \)

Case 1: Ground-to-Ground Projectile Motion

In this case, the object lands back on the same horizontal level it was projected from. The motion is symmetric.

  • Time to reach maximum height: \( t_{\text{up}} = \dfrac{u\sin\theta}{g} \)
  • Total time of flight: \( T = \dfrac{2u\sin\theta}{g} \)
  • Maximum height: \( H = \dfrac{u^2\sin^2\theta}{2g} \)
  • Horizontal range: \( R = \dfrac{u^2\sin(2\theta)}{g} \)

Example:

A ball is thrown with speed \( 25\,\text{m/s} \) at an angle \( 45^\circ \) from level ground. Calculate:

  • (a) Time of flight
  • (b) Maximum height
  • (c) Range
▶️ Answer/Explanation

Given: \( u = 25\,\text{m/s},\ \theta = 45^\circ,\ g = 9.8\,\text{m/s}^2 \)

(a) Time of Flight:

\( T = \dfrac{2u\sin\theta}{g} = \dfrac{2 \cdot 25 \cdot \sin 45^\circ}{9.8} = \dfrac{50 \cdot 0.707}{9.8} \approx \boxed{3.61\,\text{s}} \)

(b) Maximum Height:

\( H = \dfrac{u^2\sin^2\theta}{2g} = \dfrac{625 \cdot 0.5}{19.6} \approx \boxed{15.95\,\text{m}} \)

(c) Horizontal Range:

\( R = \dfrac{u^2\sin(2\theta)}{g} = \dfrac{625 \cdot 1}{9.8} \approx \boxed{63.78\,\text{m}} \)

Case 2: Projectile Motion from a Height

In this case, the projectile is launched from an elevated position or lands at a different height from the starting point.

  • The total time is found by solving: \( y = u\sin\theta \cdot t – \dfrac{1}{2}gt^2 + h_0 \)
  • The range depends on time of flight: \( R = u\cos\theta \cdot t \)
  • Maximum height is still given by: \( H = \dfrac{(u\sin\theta)^2}{2g} \) , if we launch a projectile from some initial height h then this “h” will be add in maximum height.

Example:

A stone is thrown upward at \( 15\,\text{m/s} \) from the top of a tower \( 20\,\text{m} \) high. How long does it take to hit the ground?

▶️ Answer/Explanation

Use \( s = ut + \dfrac{1}{2}gt^2 \), with downward taken as positive:
\( s = 20\,\text{m},\ u = -15\,\text{m/s},\ g = 9.8\,\text{m/s}^2 \)
\( 20 = -15t + 4.9t^2 \Rightarrow 4.9t^2 – 15t – 20 = 0 \)

Solving quadratic:

\( t = \dfrac{15 \pm \sqrt{225 + 392}}{2 \cdot 4.9} = \dfrac{15 \pm \sqrt{617}}{9.8} \approx \dfrac{15 \pm 24.82}{9.8} \)
Reject negative value → \( t \approx \boxed{4.06\,\text{s}} \)

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