An Olympic long jumper leaves the ground at an angle of 23^o and travels through the air for a horizontal distance of 8.7m before landing. The takeoff speed of the jumper is-[Assume g = 9.8ms^-2]

A. 19.8ms^-1

B. 14.7ms^-1

C. 17.3ms^-1

D. 10.9ms^-1

সঠিক উত্তরঃ D. 10.9ms^-1

Explanation:

Another Explanation (5): ```html

Olympic Long Jump Solution

Given:

We need to find the takeoff speed, \(v_0\).

The range \(R\) of a projectile is given by:

\(R = \frac{v_0^2 \sin(2\theta)}{g}\)

We can rearrange this formula to solve for \(v_0\):

\(v_0 = \sqrt{\frac{Rg}{\sin(2\theta)}}\)

Now, plug in the given values:

\(v_0 = \sqrt{\frac{8.7 \times 9.8}{\sin(2 \times 23^\circ)}}\)

\(v_0 = \sqrt{\frac{85.26}{\sin(46^\circ)}}\)

\(v_0 = \sqrt{\frac{85.26}{0.7193}}\)

\(v_0 = \sqrt{118.53}\)

\(v_0 \approx 10.9\) m/s

Therefore, the takeoff speed of the jumper is approximately \(10.9\) m/s. 🚀

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