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]

• Angle of takeoff, \(\theta = 23^\circ\)
• Horizontal distance (range), \(R = 8.7\) m
• Acceleration due to gravity, \(g = 9.8\) m/s²

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

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

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

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

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

\(v_0 = \sqrt{118.53}\)

\(v_0 \approx 10.9\) m/s