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2025 WAEC Physics Theory 1a. Explain what is meant by the acceleration of free fall due to gravity, g. b. State...

Physics
WAEC 2025

1a. Explain what is meant by the 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 100 ms\(^{-1}\), calculate the:

i. maximum height of the stone above the ground (neglect air resistance and take g as 10 ms\(^{-2}\))

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 10 ms\(^{-2}\))

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Explanation

1a. Acceleration of a free fall due to gravity is the force of attraction of the Earth on a unit mass.

b. g varies on the surface of the earth because of the shape of the earth and the rotation of the earth about its polar axis.

ci. θ = 30°; H = 100 m; u = 100 m/s; h = ?

 Total height = H + h

=> h = \(\frac{u^2 sin^2 θ }{2g}\) = \(\frac{100^2 sin^2 30°}{ 2 \times 10}\)

= \(\frac{10000 \times 0.25}{ 20}\) = \(\frac{2500}{ 20}\) = 125 m

. Height from the ground = height of tower + the maximum height the stone reached = 100 + 125 = 225 m

ii. t = \(\frac{u sin θ }{\text{g}}\) = \(\frac{100 \times sin 30°}{10}\) = \(\frac{100 \times 0.5 }{10}\) = 10 × 0.5 = 5 s

iii. Time of flight = time taken for a particle which is projected to return to its original level T = \(\frac{2u sin θ}{\text{g}}\) = \(\frac{2 \times 100 \times sin 30°}{10}\) = \(\frac{200 \times 0.5}{10}\) = 10 s

Time of flight = 10 s

But the total time taken to land on the ground = T + time used to return from the top of the tower to the ground level. Since it's falling S\(_y\) = 100 m

S\(_y\) = ut sin θ - \(\frac{1}{2}\) g t\(^2\) => -100

= 100t sin 30° -  \(\frac{1}{2}\) (10) t\(^2\)  =>

-100 = 100t × 0.5 - 5 t\(^2\) => -100 = 50t - 5 t\(^2\) => 5 t\(^2\) - 50t - 100 = 0 => t\(^2\) - 10t - 20 = 0

t = \(\frac{10 \pm \sqrt{10^2 - 4 \times -20 \times 1}}{2}\) = \(\frac{10 \pm \sqrt{100 + 80}}{2}\)

t = \(\frac{10 + 13.416}{2}\) = \(\frac{23.416}{2}\) ≈ 11.7s (ignore the -ve part)

t = 11.7 s => Total time = 10 + 11.7 = 21.7 seconds

iv. Horizontal distance = the range = \(\frac{u^2 sin 2θ}{\text{g}}\) = \(\frac{100^2 sin 60°}{10}\) = \(\frac{10000 \times 0.866}{10}\) = 1000 × 0.866 = 866 m


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