Projectile Motion (DP IB Physics: HL): Revision Note

A stone is projected from the top of a cliff 50.0 m high at an angle of 25.0° below the horizontal. The stone has an initial speed of 30.0 m s⁻¹ and follows a curved trajectory. The stone hits the ground at a horizontal distance D from the base of the cliff with a vertical velocity of 33.8 m s⁻¹.

Calculate the horizontal distance D.  

Answer:

Step 1: Understand the information given in the question

• The final vertical velocity of the stone, v = 33.8 m s⁻¹ 

• The horizontal velocity will remain constant throughout the motion

• The question wants us to calculate the range of the stone

• The vertical acceleration is +g m s⁻²

Step 2: Resolve velocity into the vertical and horizontal components

Draw a triangle on the diagram to show the vertical and horizontal velocity components

Calculate the initial vertical component of velocity, u_v using trigonometry:

• u_v = opposite side

• u_v = sinθ × hypotenuse side

• u_v = sin(25) × 30 = 12.68 m s⁻¹

Calculate the initial horizontal component of velocity, u_H using trigonometry: 

• u_H = adjacent side

• u_H = cosθ × hypotenuse side

• u_H = cos(25) × 30 = 27.19 m s⁻¹

Step 3: Consider the equations of motion in the vertical and horizontal directions

Step 4: Calculate the time of flight from the vertical motion

v = u + at

33.8 = 12.68 + 9.81t

33.8 - 12.68 = 9.81t

$t=\frac{33.8-12.68}{9.81}$

t = 2.15 s

Step 5: Calculate the range of the stone, D using the elapsed time and horizontal motion

s = ut

D = 27.19 × 2.15

D = 58.46 m = 58 m (2 s.f.)

A ball is thrown from a point P with an initial velocity u of 12 m s^-1 at 50° to the horizontal.

What is the value of the maximum height at Q? (ignoring air resistance)

Answer:

Answer:

Step 1: Consider the vertical motion and list the known quantities:

• Maximum height, $s$

• Final velocity, $v=0$

• Acceleration, $a=-9.81\mathrm{m}{\mathrm{s}}^{-2}$

Step 2: Calculate the vertical component of velocity

$\mathrm{vertical}\mathrm{velocity}=\sin \theta \times \mathrm{resultant}\mathrm{velocity}$

$u=12\sin \left(50\right)$

Step 2: Calculate the maximum height using the quantities listed in Step 1

• The equation that links $s$, $u$, $v$ and $a$ is

$v^2=u^2\hspace{0.17em}+2as$

$2as=v^2-u^2$

$s=\frac{v^2-u^2}{2a}=\frac{0-{\left(12\sin \left(50\right)\right)}^2}{2\times \left(-9.81\right)}$

$s=4.3\mathrm{m}$

A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown.

What was the speed at take-off? (ignoring air resistance)

Answer:

Step 1: Consider the vertical motion first and list the known quantities:

• Height, $s=1.25\mathrm{m}$

• Initial velocity, $u=0$

• Acceleration, $a=9.81\mathrm{m}{\mathrm{s}}^{-2}$

• Time of flight, $t$

Step 2: Calculate the time of flight using the quantities listed in Step 1

• The equation that links $s$, $u$, $a$ and $t$ is

$s=ut+\frac{1}{2}a{t}^2$

$s=\frac{1}{2}a{t}^2$

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\times 1.25}{9.81}}=0.505\mathrm{s}$

Step 3: Consider the horizontal motion and list the known quantities:

• Distance travelled, $s=10\mathrm{m}$

• Initial velocity, $u$

• Acceleration, $a=0$

• Time of flight, $t=0.505\mathrm{s}$

Step 4: Calculate the speed at take-off

• Since the acceleration is zero, use the following equation

$\mathrm{velocity}=\frac{\mathrm{displacement}}{\mathrm{time}}$

$u=\frac{10}{0.505}$

$u=19.8\mathrm{m}{\mathrm{s}}^{-1}$

Make sure you don’t make these common mistakes:

• Mixing up positive and negative values for vectors

• Mixing up velocities and distances between horizontal and vertical motion

• Confusing the direction of sin θ and cos θ

• Not converting units (mm, cm, km etc.) to metres

Furthermore, it is worth noting that projectile motion is typically symmetrical when air resistance is ignored, allowing for the use of the peak to find the total time of flight or total horizontal distance by doubling the amount to get from the start point to the peak.

In these exam questions, unless specified, fluid resistance can be ignored.