WAEC SSCE - Physics - 1994 (Objective)

Pergunta 1 Relatório

A ball is dropped from a height of 45m above the ground. Calculate the velocity of the ball just before it strikes the ground. (Neglect air resistance and take g as \(10ms^{-2}\)

21.2\(ms^{-1}\)

30\(ms^{-1}\)

300\(ms^{-1}\)

450\(ms^{-1}\)

900\(ms^{-1}\)

Detalhes da Resposta

The velocity of an object just before it strikes the ground can be found using the formula: v = \(\sqrt{2gh}\) where v is the velocity, g is the acceleration due to gravity, and h is the height from which the object was dropped. In this question, the height from which the ball was dropped is given as 45m, and the acceleration due to gravity is given as \(10ms^{-2}\). Substituting these values in the formula, we get: v = \(\sqrt{2 \times 10 \times 45}\) v = \(\sqrt{900}\) v = 30\(ms^{-1}\) Therefore, the velocity of the ball just before it strikes the ground is 30\(ms^{-1}\). So, the correct answer is the second option: 30\(ms^{-1}\).






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