(a) Long jumper’s take off speed. (b) How long or short of the opposite bank will she land? Solution: (a) Long jumper’s take off speed v 0 = 8.55 m/s (b)Horizontal distance during fly = 10 m/s .So, she just lands opposite bank of river. Explanation: Given: (a)Angle of projection = 45 0 Horizontal distance (R) = 8 (b)Angle of projection = 45 o Vertical displacement= 2.5 m downward Take off speed v o = 8.85 m/s Formula used: Range (Horizontal distance) R = u 2 sin 2 θ g Where, u is the initial velocity, θ is the projectile angle and g is the acceleration due to gravity. Second equation of motion: s = u t + 1 2 a t 2 Where, s is the displacement, t is the time, a is the acceleration and u is the initial velocity. Calculation: (a) R = u 2 sin 2 θ g 8 m = u 2 × sin ( 2 × 45 o ) 9.8 u 2 = 8 × 9.8 u = 8.85 m/s (b) Horizontal component of takeoff velocity = v o sin 45 o v x = 8.85 × 1 2 v x = 6.26 m/s Vertical component of takeoff velocity = v 0 cos 45 o = 8.85 × 1 2 v y = 6.26 m/s Apply the formula (vertical motion) s = u t + 1 2 a t 2 Take upward as negative and downward as positive s = h = 2.5 m u = v y = − 6.26 m/s a = g = 9.8 m/s 2 2.5 = − 6.26 t + 1 2 × 9.8 × t 2 4.9 t 2 − 6.26 t − 2.5 = 0 t = 6.26 ± 6.26 2 − 4 ( 4.49 ) ( − 2.5 ) 2 × 4.9 t = 1.59 s or -0 .319 s Negative value of time is not accepted Therefore, t = 1.6 s Horizontal distance travelled = v x × t = 6.25 × 1.6 = 10 m Hence, she just land at the opposite bank because horizontal distance travelled = 10 m and horizontal distance of opposite bank is also 10 m . Conclusion: Long jumper’s take off speed v 0 = 8.55 m/s . Horizontal distance during fly = 10 m/s . So, she just lands opposite bank of river.

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