State, without any calculations, how the graph could be used to determine the distance fallen.

(i) In the space below, draw and label arrows to represent the forces on the ball at 2.0 s.

(ii) Use the graph opposite to show that the acceleration of the ball at 2.0 s is approximately 4 ms⁻².

(iii) Calculate the magnitude of the force of air resistance on the ball at 2.0 s.

(iv) State and explain whether the air resistance on the ball at t = 5.0 s is smaller than, equal to or greater than the air resistance at t = 2.0 s.

After 10 s the ball has fallen 190 m.

(i) Show that the sum of the potential and kinetic energies of the ball has decreased by 780 J.

(ii) The specific heat capacity of the ball is 480 J kg⁻¹ K⁻¹. Estimate the increase in the temperature of the ball.

(iii) State an assumption made in the estimate in (c)(ii).

the area under the curve;

(i) arrows as shown, with up arrow shorter;

(ii) drawing of tangent to curve at t = 2.0 s;
calculation of slope of tangent in range 3.6 − 4.4ms⁻² ; 
Award [0] for calculations without a tangent but do not be particular about size of triangle.

(iii) calculation of F = ma = 0.50 × 4 = 2N
R(= mg −ma = 0.50×9.81− 0.50× 4) ≈ 3N;

(iv) the acceleration is decreasing;
and so R is greater;
or
air resistance forces increase with speed;
since speed at 5.0 s is greater so is resistance force;

(i) loss of potential energy is mgΔh=0.50×9.81×190=932J;
gain in kinetic energy is \(\frac{1}{2}m{v^2} = \frac{1}{2}0.50 \times {25^2} = 156{\rm{J}}\);
loss of mechanical energy is 932–156;
≈780J

(ii) mcΔθ=780J;
\(\Delta \theta  = \left( {\frac{{780}}{{0.5 \times 480}}} \right) \approx 3{\rm{K}}/{3^ \circ }{\rm{C}}\);
 
(iii) all the lost energy went into heating just the ball / no energy transferred to surroundings / the ball was heated uniformly;