In his first attempt, a long jumper took off from the spring board with a...
PHYSICS
WAEC 2004
In his first attempt, a long jumper took off from the spring board with a speed of 8 ms\(^{-1}\) at 30° to the horizontal. He makes a second attempt with the same speed at 45° to the horizontal. Given that the expression for the horizontal range of a projectile is \(\frac{u^2 sin 2\theta}{g}\) where all the symbols have their usual meanings, show that he gains a distance of 0.8576 m in his second attempt. [g = 10ms\(^{-2}\)]
Explanation
1st Attempt R = \(\frac{U^2 sin 2 \theta}{g} = \frac{8^2 sin 60}{10} = 5.5424m\)
2nd Attempt R = \(U^2sin 2 \theta\)
= \(\frac{8^2sin 90}{10}\) = 6.4m
Gain = 6.4 - 5.5425 = 0.8576m
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