A ball has (time in s, vertical position in m) data (0, 0), (0.4, 5.216), (0.8, 8.864), (1.2, 10.944), (1.6, 11.456), (2, 10.4), (2.4, 7.776) and constant horizontal velocity 9.0 m/s. (a) Use y = uᵧt − ½gt² to estimate uᵧ and g. (b) Find model maximum time and height. (c) Find time and range on return to launch level. (d) State accelerations. (e) Explain two departures from an ideal parabola.

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Accepted Answer

uᵧ = 15.0 m/s; g = 9.80 m/s²; peak 1.53 s and 11.5 m; flight 3.06 s; range 27.6 m; accelerations 0 horizontally and −9.8 m/s² vertically. For example at 0.8 s: 8.864 = 0.8uᵧ − 0.32g, and at 2.0 s: 10.4 = 2uᵧ − 2g. Solving gives uᵧ = 15.0 m/s and g = 9.80 m/s². At the peak vᵧ = uᵧ − gt = 0, so t = 15.0/9.80 = 1.53 s. Height uᵧ²/(2g) = 225/19.6 = 11.5 m, consistent with nearby samples. The non-zero solution of y = 0 is t = 2uᵧ/g = 3.06 s. Horizontal range = 9.0 × 3.06 = 27.6 m. Neglecting resistance, horizontal acceleration is zero and vertical acceleration remains −9.8 m/s², including at maximum height where only vertical velocity is momentarily zero. Air resistance reduces horizontal speed and changes symmetry. Camera scale, parallax, timing offset or tracking the ball edge instead of its centre can also move points away from the model.

Projectile trajectory reconstructed from data

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Projectile vertical position

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Projectile vertical position — supplied values
Time / s	Vertical position / m
0	0
0.4	5.216
0.8	8.864
1.2	10.944
1.6	11.456
2	10.4
2.4	7.776