NCEA Scholarship Physics free practice questions with answers

Question 1 (8 marks) [Projectile motion from a height]

(a) A stone is launched at 18.0 m s⁻¹ at 35.0° above the horizontal from the edge of a cliff, 22.0 m above level ground. Ignore air resistance. Calculate the time of flight and the horizontal distance travelled before it lands.

(a) Answer: T = 3.42 s; R = 50.4 m

• Entry step: worth marks in the 1-4 range of the 8-point question score.

• Take up as positive: initial vertical speed 18.0 sin 35.0° = 10.32 m s⁻¹ and horizontal speed 18.0 cos 35.0° = 14.74 m s⁻¹. Landing: −22.0 = 10.32 t − ½(9.81)t². [1 mark]

• Solve the quadratic and take the positive root: t = 3.417 s. Horizontal distance = 14.74 × 3.417 = 50.39 m. [1 mark]

(b) Find the speed and the angle below the horizontal of the velocity at the instant of landing. Show that this speed would be the same for any launch angle.

(b) Answer: v = 27.5 m s⁻¹ at 57.6° below the horizontal

• Scholarship band (5-6): the part needs your own method and accurate algebra.

• Vertical speed at landing = 10.32 − 9.81(3.417) = −23.20 m s⁻¹ (downwards). Speed = 27.49 m s⁻¹.

[1 mark]

• Angle = tan⁻¹(23.20 ÷ 14.74) = 57.56° below the horizontal. [1 mark]

• Energy conservation: ½mv² = ½mu² + mgh, so v = √(u² + 2gh) depends only on u and h, not on the launch angle. [1 mark]

(c) It is given that, for a launch from height h at speed u, the horizontal range is greatest when tan θ = u ÷ √(u² + 2gh). Find this angle and the greatest range for the data above, and show that a launch at 60.0° gives a shorter range.

(c) Answer: θ = 33.2°; Rmax = 50.4 m (at 60.0° the range is 38.1 m)

• Outstanding band (7-8): the part needs insight and a complete, efficient argument.

• tan θ = 18.0 ÷ 27.49 = 0.6548 so θ = 33.22°. [1 mark]

• At this angle, Rmax = u√(u² + 2gh) ÷ g = 18.0 × 27.49 ÷ 9.81 = 50.44 m. [1 mark]

• At 60.0°: T = 4.237 s and R = 18.0 × cos 60.0° × 4.237 = 38.13 m, which is less than 50.44 m. [1 mark]