WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Kinematics
Projectiles launched at an angle
- 2 things to know
- AS Unit 1
Overview
Recall questions on Projectiles launched at an angle.
What you need to know
Answer each one before you reveal it.
Why does a projectile launched over level ground take the same time to rise as it does to fall?
Questions from this subtopic
Each one with the model answer underneath.
- 1
A ball is launched at 24 m s⁻¹ at 30° above the horizontal. Calculate the horizontal and vertical components of its initial velocity.
Model answer
The horizontal component is 24 cos 30° = 20.8 m s⁻¹ and the vertical component is 24 sin 30° = 12.0 m s⁻¹ upwards.
- 2
A ball is kicked at 18 m s⁻¹ at 55° above the horizontal. Calculate the time it takes to reach its highest point, taking g as 9.81 m s⁻².
Model answer
The vertical component of the launch velocity is 18 sin 55° = 14.7 m s⁻¹, and this falls to zero in 14.7 ÷ 9.81 = 1.50 s.
- 3
An arrow is shot at 40 m s⁻¹ at 25° above the horizontal. Calculate the greatest height it reaches above its launch point, taking g as 9.81 m s⁻².
Model answer
The vertical component of the launch velocity is 40 sin 25° = 16.9 m s⁻¹, and the rise is 16.9² ÷ (2 × 9.81) = 14.6 m.
- 4
A stone is thrown at 15 m s⁻¹ at 40° above the horizontal over level ground. Calculate its total time of flight, taking g as 9.81 m s⁻².
Model answer
The vertical component of the launch velocity is 15 sin 40° = 9.64 m s⁻¹, and the flight lasts 2 × 9.64 ÷ 9.81 = 1.97 s.
- 5
A ball is struck at 22 m s⁻¹ at 35° above the horizontal over level ground. Calculate its horizontal range, taking g as 9.81 m s⁻².
Model answer
The horizontal component is 22 cos 35° = 18.0 m s⁻¹ and the flight lasts 2 × 22 sin 35° ÷ 9.81 = 2.57 s, so the range is 46.4 m.
- 6
Why is the speed of a projectile not zero at the highest point of its path?
Model answer
Only the vertical component of the velocity is zero there. The horizontal component is unchanged from launch, and it is that component which gives the projectile its speed at the top.
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