An athlete executing a long jump leaves the ground at a 30° angle and travels 5.80 m.?
An athlete executing a long jump leaves the ground at a 30° angle and travels 5.80 m.
a) What was the takeoff of his speed?
(b) If this speed were increased by just 4.0 percent, how much longer would the jump be?
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If we assume that the ground is level, we can use the range equation to determine the athlete’s initial velocity.
Range = v^2/g * sin 2θ
5.80 = v^2/9.8 * sin 60
v^2 = 5.80 * 9.8 ÷ sin 60
v = √(58.84 ÷ sin 60) ≈ 8.24278898 m/s
(b) If this speed were increased by just 4.0 percent, how much longer would the jump be?
v = 1.04 * √(58.84 ÷ sin 60)
v^2 = 1.0816 * 58.84 ÷ sin 60 = 63.641344 ÷ sin 60
Range = (63.641344 ÷ sin 60)/9.8 * sin 60 = 63.641344 ÷ 9.8 = 6.494014694 meters