Projectile Motion and Kinematics Problems Solved

Kinematics is the geometry of motion: it describes where an object is, how fast it moves and how its speed changes, without yet asking why. Projectile motion is kinematics in two dimensions, and it appears everywhere from cricket balls and water jets to the design of conveyor discharge chutes. In this guide we build the key equations and then solve three realistic problems with every step and unit shown.

The Constant-Acceleration Equations

When acceleration \( a \) is constant, five quantities describe straight-line motion: initial velocity \( u \), final velocity \( v \), acceleration \( a \), time \( t \) and displacement \( s \). Students in Pakistan often call them the SUVAT equations:

$$ v = u + at $$

$$ s = ut + \tfrac{1}{2}at^2 $$

$$ v^2 = u^2 + 2as $$

$$ s = \tfrac{1}{2}(u + v)t $$

Each equation leaves out one of the five quantities, so you choose the one that omits whatever you neither know nor need.

A problem-solving routine

1. Draw a sketch and choose a positive direction.
2. List the knowns and the unknown with units.
3. Pick the equation that links them.
4. Solve symbolically first, then substitute numbers.
5. Check the sign, size and units of the answer.

Projectile Motion: Two Independent Motions

A projectile moving near Earth's surface with air resistance ignored has:

• Horizontal motion with zero acceleration, so \( v_x = u\cos\theta \) stays constant.
• Vertical motion with constant acceleration \( g = 9.81 \) m/s² downward.

The two motions share only one thing: time. That insight solves almost every projectile problem.

For a launch from level ground at speed \( u \) and angle \( \theta \), these results follow:

$$ T = \frac{2u\sin\theta}{g}, \qquad H = \frac{u^2\sin^2\theta}{2g}, \qquad R = \frac{u^2\sin 2\theta}{g} $$

where \( T \) is time of flight, \( H \) maximum height and \( R \) horizontal range. Because \( \sin 2\theta \) peaks at \( 2\theta = 90^\circ \), the maximum range on level ground occurs at \( \theta = 45^\circ \), and complementary angles such as 30° and 60° give equal ranges.

Solved Problem 1: Braking Distance on a Motorway

A car travels at 72 km/h. The driver reacts after 1.2 s and then brakes with a constant deceleration of 6.0 m/s². Find the total stopping distance.

Step 1: Convert units.

$$ u = 72\ \frac{\text{km}}{\text{h}} \times \frac{1000\ \text{m}}{3600\ \text{s}} = 20\ \text{m/s} $$

Step 2: Reaction distance. The car still moves at constant speed during the reaction time:

$$ s_1 = ut = (20\ \text{m/s})(1.2\ \text{s}) = 24\ \text{m} $$

Step 3: Braking distance. Use \( v^2 = u^2 + 2as \) with \( v = 0 \) and \( a = -6.0 \) m/s²:

$$ s_2 = \frac{0 - u^2}{2a} = \frac{-(20\ \text{m/s})^2}{2(-6.0\ \text{m/s}^2)} = \frac{400}{12}\ \text{m} = 33.3\ \text{m} $$

Step 4: Total.

$$ s = 24 + 33.3 = 57.3\ \text{m} $$

Sanity check. The braking time is \( 20/6 = 3.33 \) s, and the average speed while braking is 10 m/s, so \( 10 \times 3.33 = 33.3 \) m, confirming Step 3. Notice that braking distance grows with the square of speed: at 144 km/h it would be four times longer, 133 m, which is why speed limits matter so much.

Solved Problem 2: A Ball Launched From Level Ground

A ball is launched at 25 m/s at 40° above the horizontal on level ground. Find the time of flight, the maximum height and the range.

Step 1: Resolve the velocity.

$$ u_x = 25\cos 40^\circ = 25(0.7660) = 19.15\ \text{m/s} $$

$$ u_y = 25\sin 40^\circ = 25(0.6428) = 16.07\ \text{m/s} $$

Step 2: Time of flight. The ball returns to launch height when vertical displacement is zero:

$$ 0 = u_y t - \tfrac{1}{2}gt^2 \quad \Rightarrow \quad t = \frac{2u_y}{g} = \frac{2(16.07\ \text{m/s})}{9.81\ \text{m/s}^2} = 3.28\ \text{s} $$

Step 3: Maximum height. At the top \( v_y = 0 \):

$$ H = \frac{u_y^2}{2g} = \frac{(16.07)^2}{2(9.81)}\ \text{m} = \frac{258.2}{19.62}\ \text{m} = 13.2\ \text{m} $$

Step 4: Range.

$$ R = u_x T = (19.15\ \text{m/s})(3.276\ \text{s}) = 62.7\ \text{m} $$

Sanity check. Using the formula directly, \( R = u^2\sin 80^\circ / g = (625)(0.9848)/9.81 = 62.7 \) m, which agrees. At 45° the range would be \( 625/9.81 = 63.7 \) m, slightly more, as expected because 40° is close to the optimum.

Solved Problem 3: Material Leaving a Raised Conveyor

Gravel leaves the end of a horizontal conveyor at 12 m/s from a height of 45 m above the ground of a quarry. Where does it land, and how fast is it moving on impact?

Step 1: Vertical motion gives time. Initial vertical velocity is zero:

$$ 45 = \tfrac{1}{2}(9.81)t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2(45\ \text{m})}{9.81\ \text{m/s}^2}} = \sqrt{9.174\ \text{s}^2} = 3.03\ \text{s} $$

Step 2: Horizontal distance.

$$ x = (12\ \text{m/s})(3.029\ \text{s}) = 36.3\ \text{m} $$

Step 3: Velocity components at impact.

$$ v_x = 12\ \text{m/s}, \qquad v_y = gt = (9.81)(3.029) = 29.7\ \text{m/s} $$

Step 4: Impact speed and angle.

$$ v = \sqrt{12^2 + 29.7^2} = \sqrt{144 + 882.9} = 32.0\ \text{m/s} $$

$$ \theta = \tan^{-1}\left(\frac{29.7}{12}\right) = 68.0^\circ \text{ below the horizontal} $$

Sanity check. Energy conservation gives \( v = \sqrt{u^2 + 2gh} = \sqrt{144 + 882.9} = 32.0 \) m/s, the same answer by a completely different route. An engineer would place the stockpile centre roughly 36 m from the conveyor end.

Common Errors to Avoid

• Forgetting km/h to m/s conversion. Divide by 3.6.
• Using the level-ground range formula when launch and landing heights differ, as in Problem 3.
• Mixing signs. If up is positive, then \( g \) enters as \( -9.81 \) m/s².
• Setting \( v = 0 \) at the top for the whole velocity. Only the vertical component is zero; the horizontal component continues.
• Calculator in radian mode when angles are in degrees.

Did You Know?

• Galileo showed around 1638, in Two New Sciences, that a projectile's path is a parabola by treating horizontal and vertical motion separately.
• Air resistance is far from negligible for fast, light objects: a well-struck cricket ball travels noticeably less far than the ideal formula predicts.
• On the Moon, where \( g \) is about 1.62 m/s², the same throw would travel about six times as far.

Key Takeaways

• The SUVAT equations hold only for constant acceleration; pick the one that omits the quantity you do not need.
• Split projectile motion into independent horizontal (constant velocity) and vertical (constant acceleration) parts linked by time.
• Level-ground formulas for \( T \), \( H \) and \( R \) are shortcuts; for unequal heights, work from first principles.
• Verify answers with energy conservation or an alternative formula.

If kinematics or dynamics assignments are slowing you down, Amathify's subject experts and PhD scholars can provide clear, step-by-step solutions; contact us on WhatsApp or through the Ask a Question form.