2. The equation that solves a problem is 6.4 m-20 m + 3.0 m (2.0 s)-49 (20 s). T

Question

2. The equation that solves a problem is 6.4 m-20 m + 3.0 m (2.0 s)-49 (20 s). The problem is: a. How far above its initial position does a rock travel in 2.0 s when thrown b. How far below its initial position does a rock travel in 2.0 s when thrown c. What is the position relative to the ground of a rock thrown up at 3.0 m/s d. What is the change in position relative to the ground of a rock thrown up at e. What is the position relative to the ground of a rock thrown up at 3.0 m/s up from a point 40 m above the ground? up from a point 40 m above the ground? from a roof 20 m above the ground 2.0 s after it is released? 3.0 m/s from a roof 20 m above the ground 2.0 s after it is released? from a roof 20 m above the ground if its maximum height is 33.6 m?

Explanation / Answer

Used equation is similar to

h2 = h1 + u*t + 0.5*a*t^2

where u = initial speed = 3.0 m/sec

time = 3 sec

a = -9.8

h2 - h1 = 6.4 - 20 = 13.6 m

Option A and B is wrong, Since initial velocity is not given

Option C & D is wrong because, h2 - h1 = 20 m

Option E is Correct.

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