After 21 seconds the wheel turns through 21/30×360=252°. Let the centre of the wheel be O and position on the wheel’s circumference after 21 seconds be Q.
The starting point is 3 o’clock position which is 90° from the lowest position (P) on the circumference of the wheel. 90+252=342° which makes Q 360-342=18° clockwise from P. Let QN be the perpendicular from Q onto OP. OQ=5in, ON=5cos18, therefore PN=5-5cos18 which is 6.5-5cos18 from the ground and is the same as the height of Q from the ground=1.74in approx.