3.3.5 Solve simple projectile problems (A-level only)

Multiple choice

In the projectile model, horizontal acceleration is $[...]\text{[...]}$ and vertical acceleration is $- g$ when upward is positive.

For launch speed $u$ at angle $θ\theta$ above the horizontal, $u_x=$ $ucos⁡θu\cos\theta$ and $u_y=$ $usin⁡θ\boxed{u\sin\theta}$ .

In the projectile model, horizontal acceleration is $0\boxed{0}$ and vertical acceleration is $- g$ when upward is positive.

For launch speed $u$ at angle $θ\theta$ above the horizontal, $u_x=$ $ucos⁡θ\boxed{u\cos\theta}$ and $u_y=$ $usin⁡θu\sin\theta$ .

$v =$ $vx2+vy2\sqrt{v_x^2+v_y^2}$ .

Multiple choice

In the projectile model, horizontal acceleration is $[...]\text{[...]}$ and vertical acceleration is $- g$ when upward is positive.

For launch speed $u$ at angle $θ\theta$ above the horizontal, $u_x=$ $ucos⁡θu\cos\theta$ and $u_y=$ $usin⁡θ\boxed{u\sin\theta}$ .

In the projectile model, horizontal acceleration is $0\boxed{0}$ and vertical acceleration is $- g$ when upward is positive.

For launch speed $u$ at angle $θ\theta$ above the horizontal, $u_x=$ $ucos⁡θ\boxed{u\cos\theta}$ and $u_y=$ $usin⁡θu\sin\theta$ .

$v =$ $vx2+vy2\sqrt{v_x^2+v_y^2}$ .

3.3.5 Solve simple projectile problems (A-level only) Flashcards