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A projectile is launched from the top of a cliff of height h = 10 m above the ground at an angle theta above the horizontal. Exactly t1 = 1 s after launch, the projectile passes the level of the cliff top again while moving downward. It eventually lands on the ground at a horizontal distance d = 10 m from the launch point. Find the value of 2*tan(theta).

1. 1
2. 2
3. 3
4. 4

Correct answer: 2

Solution

At t1=1s the projectile is back at cliff-top level: vy*1 - (1/2)*g*1² = 0 gives vy = 5 m/s. Total flight time from ground landing: y = -10 gives t = 2s, so vx = 10/2 = 5 m/s. Thus tan(theta) = vy/vx = 1 and 2*tan(theta) = 2.

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