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Answer:
time: 1.122 secondsrange: 10.693 mmaximum height: 1.543 mExplanation:
Given:
runner is launched at 30° angle to horizontal at 11 m/s
acceleration due to gravity is g = -9.8 m/s²
Find:
runner's hang time
runner's distance to the landing point
runner's maximum height
Solution:
The (horizontal, vertical) speed components will be ...
(11 m/s)(cos(30°), sin(30°)) = (5.5√3 m/s, 5.5 m/s)
The time of flight can be found from the height formula:
h(t) = 1/2gt² +vt . . . . . . where v is the vertical speed at launch
The time we're concerned with is the time when h(t)=0 and t>0.
0 = -4.9t^2 +5.5√3t = t(-4.9t +5.5√3)
The second factor is zero when ...
t = (5.5√3)/4.9 ≈ 1.122 . . . seconds hang time
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The distance to the landing point will be the product of horizontal speed and hang time:
d = (5.5 m/s)(5.5√3/4.9 s) ≈ 10.693 m . . . . distance to landing
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The maximum height can be found from the formula (based on conversion of kinetic energy to potential energy) ...
h = v²/|2g| = (5.5 m/s)²/(2(9.8 m/s²)) ≈ 1.543 m . . . . maximum height
