Physics 9702 · AS & A Level · Equations of motion

Equations of motion — practice question

A golfer plays a shot so that the ball leaves level ground with a speed of $6.0\,\text{m s}^{-1}$ at an angle $\theta$ to the horizontal, as shown in Fig. 1.1. The size of the initial vertical component $v_y$ of the velocity is $4.8\,\text{m s}^{-1}$. Take air resistance to be negligible.
(a)[1]

Show that the size of the initial horizontal velocity component $v_x$ is $3.6\,\text{m s}^{-1}$.

(b(i))[2]

On Fig. 1.2, sketch a line to illustrate how the vertical component $v_Y$ of the velocity changes with time $t$ until the ball returns to the ground (label the line Y).

(b(ii))[2]

On Fig. 1.2, sketch a line to illustrate how the horizontal component $v_X$ of the velocity varies with time $t$ until the ball returns to the ground (label the line X).

(c)[2]

Calculate the maximum height reached by the ball.

(d)[4]

For the ball’s motion from the ground to its maximum height, determine the ratio $$\frac{\text{kinetic energy at maximum height}}{\text{change in gravitational potential energy}}.$$

(e)[1]

In reality, the ball experiences significant air resistance. Explain why the true time taken for the ball to reach maximum height is shorter than the time found when air resistance is treated as negligible.

Worked solution & mark scheme

This 12-mark question has a full step-by-step worked solution and mark scheme. One marking point: $v_x=\sqrt{6.0^2-4.8^2}=3.6\ \text{m s}^{-1}$, or $6.0\sin\theta=4.8$ so $\theta=53.1^\circ$, followed by $v_x=6.0\cos53.1^\circ=3.6\ \text{m s}^{-1}$

  • Full mark scheme, point by point
  • Step-by-step worked solution
  • Write your answer & get it marked instantly by AI