(a) Explain what is meant by acceleration of free fall due to gravity, g. (b)...
(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