(a) Explain what is meant by acceleration of free fall due to gravity, g. (b) State two reasons why g varies on the surface of the earth (c) A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower of height 100 m and it hits the ground at a point Q. If the initial velocity of projection is 100ms\(^{-1}\), calculate the (i) maximum height of the stone above the ground; (ii) time it takes to reach this height; (iii) time of flight (iv) horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m\(^{-2}\))
Explanation
(a) This is the force of attraction of the Earth on a mass It is a vector quantity and is measured in ms\(^{-2}\) It pulls the acceleration when a body is in vacuum or without resistance. OR/F = g (all symbols explained) where F = force of gravity, M = mass of the object
(b) — (Shape of the earth). The earth is not a perfect sphere (1) or the shape of the earth is elliptical - Rotation of the Earth about its polar axis. (c)(i) Let H = height of tower and h = vertical difference covered by the stone from the point of projection. (i) V\(_2\) = (Usin\(\theta\))\(^2\) - 2gh \(\theta\) = 100 x 100 x (\(\frac{1}{2}\))\(^2\) - 20h h = \(\frac{2500}{20}\) = 125m
height above the ground (i) S = H + 125 S = 100 + 125 S = 225m (ii) V = U sin \(\theta\) – gt 0 = 100 sin 30° – 10t t = \(\frac{100 \times 0.5}{10}\) t = 5s (iii) Using S = ut + 1/2 gt\(^2\) 225 = 0 + 1/2 x 10t\(^2\) t = \(\sqrt{45}\) = 6.71 Time of flight, T= 5 + 6. T = 11.71s
(iv) Range R = UTcos\(\theta\) Or S = ut – 1/2 gt\(^2\) g = 0 =100 x 0.8660 x 11.71 = 1014.08 (C14) Range \(\frac{U^2 sin20}{ g}\) = 1013.22m
