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 th...
Question 1 Report
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\(^{-2}\)]
Answer Details
Using \(R = \dfrac{u^2 \sin 2\theta}{g}\) with \(u = 8\ \text{ms}^{-1}\) and \(g = 10\ \text{ms}^{-2}\), so \(u^2 = 64\).
First attempt (\(\theta = 30^{\circ}\), \(2\theta = 60^{\circ}\), \(\sin 60^{\circ} = 0.8660\)):
\[ R_1 = \frac{64 \times 0.8660}{10} = \frac{55.425}{10} = 5.5425\ \text{m} \]Second attempt (\(\theta = 45^{\circ}\), \(2\theta = 90^{\circ}\), \(\sin 90^{\circ} = 1\)):
\[ R_2 = \frac{64 \times 1}{10} = 6.4\ \text{m} \]Gain in distance:
\[ R_2 - R_1 = 6.4 - 5.5425 = 0.8575 \approx 0.8576\ \text{m} \]Hence he gains about \(0.8576\ \text{m}\) in the second attempt, as required.
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