Important actions when tackling a projectiles question

Step 1: Resolve initial velocity into components

Consider a projectile launched with initial velocity \(v_0\) at angle \(\theta\) above the horizontal.

Horizontal component: \(v_{0x} = v_0 \cos\theta\)

Vertical component: \(v_{0y} = v_0 \sin\theta\)

Step 2: Horizontal velocity (no horizontal acceleration)

Since there is no horizontal acceleration (\(a_x = 0\)), the horizontal velocity remains constant throughout the flight:

\[v_x = v_{0x} = v_0 \cos\theta\]

This remains the same at all times \(t\).

Step 3: Vertical velocity (constant downward acceleration)

The vertical motion experiences constant downward acceleration due to gravity (\(a_y = -g\)).

Using the equation: \(v = u + at\)

\[v_y = v_{0y} + a_y t\]

\[v_y = v_0 \sin\theta + (-g)t\]

\[v_y = v_0 \sin\theta - gt\]

Key insight: The horizontal velocity is constant (independent of time), while the vertical velocity decreases linearly with time due to gravity. At maximum height, \(v_y = 0\).

Step 1: Set up the problem

Consider a projectile launched from the origin with initial velocity \(v_0\) at angle \(\theta\) above the horizontal. The projectile lands at the same height from which it was launched.

Step 2: Resolve initial velocity into components

Horizontal component: \(v_{0x} = v_0 \cos\theta\)

Vertical component: \(v_{0y} = v_0 \sin\theta\)

Step 3: Find the time of flight

Using the vertical motion equation with \(s_y = 0\) (returns to same height):

\[s_y = v_{0y}t - \frac{1}{2}gt^2\]

\[0 = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2\]

\[0 = t(v_0 \sin\theta - \frac{1}{2}gt)\]

This gives \(t = 0\) (launch) or \(t = \frac{2v_0 \sin\theta}{g}\) (landing)

Therefore, time of flight: \(t = \frac{2v_0 \sin\theta}{g}\)

Step 4: Calculate the horizontal range

Since horizontal velocity is constant:

\[R = v_{0x} \times t\]

\[R = v_0 \cos\theta \times \frac{2v_0 \sin\theta}{g}\]

\[R = \frac{2v_0^2 \cos\theta \sin\theta}{g}\]

Step 5: Simplify using trigonometric identity

Using the double angle formula: \(\sin(2\theta) = 2\sin\theta\cos\theta\)

\[R = \frac{v_0^2 \times 2\sin\theta\cos\theta}{g}\]

\[R = \frac{v_0^2 \sin(2\theta)}{g}\]

Key insight: The range is maximum when \(\sin(2\theta) = 1\), which occurs at \(\theta = 45°\).

Step 1: Set up the problem

Consider a projectile launched with initial velocity \(v_0\) at angle \(\theta\) above the horizontal. We need to find the maximum vertical displacement.

Step 2: Resolve initial velocity into components

Horizontal component: \(v_{0x} = v_0 \cos\theta\)

Vertical component: \(v_{0y} = v_0 \sin\theta\)

Step 3: Key observation at maximum height

At maximum height, the vertical velocity is zero: \(v_y = 0\)

The projectile momentarily stops moving vertically before falling back down.

Step 4: Apply the vertical motion equation

Using \(v_y^2 = v_{0y}^2 - 2g s_y\) where \(s_y = H\) (maximum height):

\[0^2 = (v_0 \sin\theta)^2 - 2gH\]

\[0 = v_0^2 \sin^2\theta - 2gH\]

Step 5: Solve for H

\[2gH = v_0^2 \sin^2\theta\]

\[H = \frac{v_0^2 \sin^2\theta}{2g}\]

Key insight: Maximum height is achieved when \(\sin^2\theta = 1\), which occurs at \(\theta = 90°\) (vertical launch).

Step 1: Set up the problem

Consider a projectile launched with initial velocity \(v_0\) at angle \(\theta\) above the horizontal. The projectile lands at the same height from which it was launched.

Step 2: Resolve initial velocity into components

Horizontal component: \(v_{0x} = v_0 \cos\theta\)

Vertical component: \(v_{0y} = v_0 \sin\theta\)

Step 3: Key observation for time of flight

When the projectile returns to its launch height, the vertical displacement \(s_y = 0\).

Step 4: Apply the vertical motion equation

Using \(s_y = v_{0y}t - \frac{1}{2}gt^2\) with \(s_y = 0\):

\[0 = v_{0y}t - \frac{1}{2}gt^2\]

\[0 = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2\]

\[0 = t(v_0 \sin\theta - \frac{1}{2}gt)\]

Step 5: Solve for t

This equation gives two solutions:

\(t = 0\) (the launch time)

or

\[v_0 \sin\theta - \frac{1}{2}gt = 0\]

\[v_0 \sin\theta = \frac{1}{2}gt\]

\[t = \frac{2v_0 \sin\theta}{g}\]

Key insight: Time of flight is maximum when \(\sin\theta = 1\), which occurs at \(\theta = 90°\) (vertical launch). Note that the time of flight is twice the time to reach maximum height.

Step 1: Write parametric equations

The horizontal and vertical positions are given by:

\[x = v_0 \cos\theta \cdot t\]

\[y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2\]

These give position as a function of time \(t\).

Step 2: Eliminate time from the equations

From the horizontal equation, solve for \(t\):

\[t = \frac{x}{v_0 \cos\theta}\]

Step 3: Substitute into vertical equation

Replace \(t\) in the vertical equation:

\[y = v_0 \sin\theta \left(\frac{x}{v_0 \cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{v_0 \cos\theta}\right)^2\]

\[y = x \frac{\sin\theta}{\cos\theta} - \frac{gx^2}{2v_0^2\cos^2\theta}\]

Step 4: Simplify using trigonometry

Using \(\tan\theta = \frac{\sin\theta}{\cos\theta}\):

\[y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}\]

This is the trajectory equation in standard form.

Alternative form:

Using \(\sec^2\theta = 1 + \tan^2\theta\) and \(\cos^2\theta = \frac{1}{\sec^2\theta}\):

\[y = x\tan\theta - \frac{gx^2}{2v_0^2}(1 + \tan^2\theta)\]

This form is useful for solving problems involving specific targets.

Key insight: This equation is in the form \(y = ax + bx^2\), which is a parabola. The trajectory of any projectile is parabolic. The equation relates vertical and horizontal positions without needing to know the time.

Problem statement:

A ball is thrown horizontally with initial velocity \(v_0\) from a cliff of height \(h\). Find:

(a) The time it takes to reach the ground

(b) The horizontal distance traveled (range)

(c) The velocity when it hits the ground

Step 1: Set up coordinate system and initial conditions

Place origin at the base of the cliff, with x-axis horizontal and y-axis vertical (upward positive).

Initial position: \((0, h)\)

Initial velocity components: \(v_{0x} = v_0\), \(v_{0y} = 0\) (thrown horizontally)

Acceleration: \(a_x = 0\), \(a_y = -g\)

Step 2: Find time to reach the ground (part a)

Using vertical motion equation: \(y = y_0 + v_{0y}t + \frac{1}{2}a_y t^2\)

When the ball hits the ground, \(y = 0\):

\[0 = h + 0 \cdot t + \frac{1}{2}(-g)t^2\]

\[0 = h - \frac{1}{2}gt^2\]

\[\frac{1}{2}gt^2 = h\]

\[t = \sqrt{\frac{2h}{g}}\]

Step 3: Find horizontal distance (range) (part b)

Since horizontal velocity is constant:

\[x = v_{0x} \times t\]

\[x = v_0 \times \sqrt{\frac{2h}{g}}\]

\[x = v_0\sqrt{\frac{2h}{g}}\]

Step 4: Find velocity on impact (part c)

Horizontal velocity component (constant): \(v_x = v_0\)

Vertical velocity component using \(v = u + at\):

\[v_y = 0 + (-g)t = -g\sqrt{\frac{2h}{g}} = -\sqrt{2gh}\]

Magnitude of velocity: \[v = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + 2gh}\]

Direction below horizontal: \[\tan\phi = \frac{|v_y|}{v_x} = \frac{\sqrt{2gh}}{v_0}\]

Key insight: For horizontal projectiles, the time of flight depends only on the height (not the horizontal velocity), while the range depends on both height and initial velocity.

Problem statement:

A projectile is launched from ground level with initial velocity \(v_0 = 20\) m/s at an angle \(\theta = 30°\) above the horizontal. Find:

(a) The maximum height reached

(b) The total time of flight

(c) The horizontal range

(Take \(g = 10\) m/s²)

Step 1: Resolve initial velocity into components

Horizontal component: \(v_{0x} = v_0 \cos\theta = 20 \cos 30° = 20 \times 0.866 = 17.32\) m/s

Vertical component: \(v_{0y} = v_0 \sin\theta = 20 \sin 30° = 20 \times 0.5 = 10\) m/s

Step 2: Find maximum height (part a)

At maximum height, \(v_y = 0\). Using \(v_y^2 = v_{0y}^2 - 2gh_{max}\):

\[0 = (10)^2 - 2(10)h_{max}\]

\[20h_{max} = 100\]

\[h_{max} = 5\text{ m}\]

Step 3: Find time of flight (part b)

Using vertical motion: \(s_y = v_{0y}t - \frac{1}{2}gt^2\)

When projectile returns to ground, \(s_y = 0\):

\[0 = 10t - \frac{1}{2}(10)t^2\]

\[0 = 10t - 5t^2\]

\[0 = t(10 - 5t)\]

\(t = 0\) (launch) or \(t = 2\) s (landing)

Total time of flight: \(t = 2\) s

Step 4: Find horizontal range (part c)

Horizontal distance: \(R = v_{0x} \times t\)

\[R = 17.32 \times 2 = 34.64\text{ m}\]

Alternative using formula: \(R = \frac{v_0^2 \sin 2\theta}{g} = \frac{400 \times \sin 60°}{10} = \frac{400 \times 0.866}{10} = 34.64\) m

Summary of results:

Maximum height: 5 m

Time of flight: 2 s

Range: 34.64 m

Problem statement:

A projectile is launched from the origin with initial velocity \(v_0 = 25\) m/s. A target is located at horizontal distance \(x = 50\) m and height \(y = 10\) m. Find the launch angle \(\theta\) required to hit the target.

(Take \(g = 10\) m/s²)

Step 1: Write the trajectory equation

The general trajectory equation is:

\[y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}\]

Using the identity \(\sec^2\theta = 1 + \tan^2\theta\):

\[y = x\tan\theta - \frac{gx^2}{2v_0^2}(1 + \tan^2\theta)\]

Step 2: Substitute known values

Let \(T = \tan\theta\). Substituting \(x = 50\), \(y = 10\), \(v_0 = 25\), \(g = 10\):

\[10 = 50T - \frac{10 \times 50^2}{2 \times 625}(1 + T^2)\]

\[10 = 50T - \frac{25000}{1250}(1 + T^2)\]

\[10 = 50T - 20(1 + T^2)\]

\[10 = 50T - 20 - 20T^2\]

Step 3: Solve the quadratic equation

Rearranging:

\[20T^2 - 50T + 30 = 0\]

\[2T^2 - 5T + 3 = 0\]

Using quadratic formula: \(T = \frac{5 \pm \sqrt{25 - 24}}{4} = \frac{5 \pm 1}{4}\)

\(T = 1.5\) or \(T = 1\)

Step 4: Find the launch angles

For \(T = \tan\theta = 1.5\): \(\theta = \arctan(1.5) = 56.3°\)

For \(T = \tan\theta = 1\): \(\theta = \arctan(1) = 45°\)

Both angles will hit the target! The smaller angle (45°) gives a flatter trajectory, while the larger angle (56.3°) gives a higher, more arched trajectory.

Key insight: For most targets within range, there are typically two possible launch angles: one lower and flatter, one higher and more arched.

Problem statement:

A projectile is launched up a slope that makes an angle \(\alpha = 30°\) with the horizontal. The projectile is launched with speed \(v_0 = 20\) m/s at angle \(\beta = 45°\) to the horizontal. Find the range along the slope.

(Take \(g = 10\) m/s²)

Step 1: Set up coordinate system

Use standard horizontal-vertical coordinates with origin at launch point.

Initial velocity components:

\(v_{0x} = v_0 \cos\beta = 20 \cos 45° = 14.14\) m/s

\(v_{0y} = v_0 \sin\beta = 20 \sin 45° = 14.14\) m/s

Step 2: Equation of the slope

The slope passes through the origin with angle \(\alpha\):

\[y = x\tan\alpha = x\tan 30° = \frac{x}{\sqrt{3}}\]

Step 3: Trajectory equation of projectile

\[y = x\tan\beta - \frac{gx^2}{2v_0^2\cos^2\beta}\]

\[y = x(1) - \frac{10x^2}{2(400)(0.5)}\]

\[y = x - \frac{x^2}{40}\]

Step 4: Find intersection point

Where trajectory meets slope:

\[\frac{x}{\sqrt{3}} = x - \frac{x^2}{40}\]

\[\frac{x^2}{40} = x - \frac{x}{\sqrt{3}}\]

\[\frac{x^2}{40} = x\left(1 - \frac{1}{\sqrt{3}}\right)\]

For \(x \neq 0\):

\[\frac{x}{40} = 1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}\]

\[x = 40 \times \frac{\sqrt{3} - 1}{\sqrt{3}} = 40 \times 0.423 = 16.9\text{ m}\]

Step 5: Find range along slope

Range along slope \(R = \frac{x}{\cos\alpha} = \frac{16.9}{\cos 30°} = \frac{16.9}{0.866} = 19.5\) m

Key insight: When dealing with inclined planes, it's often easier to work in standard horizontal-vertical coordinates and then convert the final answer to slope coordinates.