Projectile Motion: Find Speed & Displacement in 2s

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SUMMARY

The discussion focuses on calculating the speed and displacement of a rock launched horizontally from a cliff at an initial velocity of 5 m/s after 2 seconds. Using the simplified gravitational acceleration of 10 m/s², the rock's horizontal displacement can be determined using the formula x = Vx * t, resulting in a distance of 10 meters from the cliff edge. The vertical displacement is calculated using the equation y = yi + Vyi * Δt + 1/2 * ay * Δt², yielding a vertical drop of 20 meters after 2 seconds.

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Homework Statement




At time t = 0 s, a large rock is launched horizontally from a tall cliff at 5 m/s. Define a coordinate system with its origin located at the edge of the cliff, its x-axis aligned with the initial velocity of the rock, and its y-axis pointing upwards. Take the acceleration of gravity to be exactly 10m/s² (rather than 9.8 m/s²) to simplify the mathematics.

a) How fast is the rock traveling and how many meters away from the edge of the cliff is the rock after 2 seconds?

b) Complete the attached table (check the attached file)

Homework Equations



y=yi + Vyi △t +1/2ay △t²

The Attempt at a Solution

 

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