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ProjectilesAQA A-Level Maths: Flashcards

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What are the modelling assumptions for a projectile?

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What are the modelling assumptions for a projectile?
Particle, no air resistance, constant acceleration gg downwards, motion in a vertical plane.
Resolve a launch speed UU at angle θ\theta above horizontal.
ux=Ucos⁡θu_x=U\cos\theta and uy=Usin⁡θu_y=U\sin\theta.
What is the horizontal acceleration of a projectile?
Zero, so vxv_x is constant.
What is the vertical acceleration of a projectile (up positive)?
−g=−9.8 m s−2-g=-9.8\ \text{m s}^{-2}.
Vector form of position under constant acceleration?
r=ut+12at2\mathbf{r}=\mathbf{u}t+\frac12\mathbf{a}t^2 with a=−gj\mathbf{a}=-g\mathbf{j}.
Condition for the highest point?
vy=0v_y=0, so the velocity is horizontal.
Greatest height formula (launch from ground)?
H=U2sin⁡2θ2gH=\frac{U^2\sin^2\theta}{2g}.
Time of flight formula (same level launch and landing)?
T=2Usin⁡θgT=\frac{2U\sin\theta}{g}.
Range formula (same level)?
R=U2sin⁡2θgR=\frac{U^2\sin2\theta}{g}.
Speed at a point where components are vxv_x and vyv_y?
vx2+vy2\sqrt{v_x^2+v_y^2}.
Equation of the path of a projectile?
y=xtan⁡θ−gx22U2cos⁡2θy=x\tan\theta-\frac{gx^2}{2U^2\cos^2\theta}, a parabola.
Stone thrown horizontally from height hh: time to land?
h=12gt2h=\frac12gt^2, so t=2hgt=\sqrt{\frac{2h}{g}}.
How does air resistance change real motion compared with the model?
Range and greatest height are reduced, and the path is no longer symmetric.

Exam questions on Projectiles

  1. A ball is projected from a point OO on level ground with speed 20 m s−120\ \text{m s}^{-1} at 30∘30^\circ above the horizontal. Model the ball as a particle moving freely under gravity, with g=9.8 m s−2g=9.8\ \text{m s}^{-2}.
    Calculate the horizontal distance the ball travels before landing.2 marks
  2. A stone is thrown horizontally at 15 m s−115\ \text{m s}^{-1} from a window 19.619.6 m above level ground. Model the stone as a particle moving freely under gravity, with g=9.8 m s−2g=9.8\ \text{m s}^{-2}.
    Find the speed of the stone as it hits the ground.2 marks
  3. A particle is projected from a point OO on horizontal ground with initial velocity (6i+14.7j) m s−1(6\mathbf{i}+14.7\mathbf{j})\ \text{m s}^{-1}, where i\mathbf{i} and j\mathbf{j} are horizontal and vertically upward unit vectors. The particle moves freely under gravity, with g=9.8 m s−2g=9.8\ \text{m s}^{-2}.
    Find the velocity of the particle after 22 s, and state whether it is moving upwards or downwards at that time.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).