The equation that solves a problem is $$6.4 \mathrm {~m} = 20 \mathrm {~m} + 3.0 \frac { \mathrm { m } } { \mathrm { s } } ( 2.0 \mathrm {~s} ) - 4.9 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } ( 2.0 \mathrm {~s} ) ^ { 2 }$$ .The problem is:
A)How far above its initial position does a rock travel in 2.0 s when thrown up from a point 40 m above the ground?
B)How far below its initial position does a rock travel in 2.0 s when thrown up from a point 40 m above the ground?
C)What is the position relative to the ground of a rock thrown up at 3.0 m/s from a roof 20 m above the ground 2.0 s after it is released?
D)What is the change in position relative to the ground of a rock thrown up at 3.0 m/s from a roof 20 m above the ground 2.0 s after it is released?
E)What is the position relative to the ground of a rock thrown up at 3.0 m/s from a roof 20 m above the ground if its maximum height is 33.6 m?

Question

Quizplus · Accepted Answer

The equation given in the problem is the equation for vertical displacement of an object (in this case, a rock) thrown upwards with an initial velocity of 3.0 m/s from a height of 40 m above ground level.

To solve for the position of the rock relative to the ground 2.0 s after it is released, we can simply substitute the values given in the problem into the equation and solve for the vertical displacement.

$$6.4 \mathrm {~m} = 20 \mathrm {~m} + 3.0 \frac { \mathrm { m } } { \mathrm { s } } ( 2.0 \mathrm {~s} ) - 4.9 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } } ( 2.0 \mathrm {~s} ) ^ { 2 }$$

$$6.4 \mathrm {~m} = 20 \mathrm {~m} + 6.0 \mathrm {~m} - 19.6 \mathrm {~m}$$

$$6.4 \mathrm {~m} = 6.4 \mathrm {~m}$$

Therefore, the position of the rock relative to the ground 2.0 s after it is released is 6.4 m above the ground.

The Equation That Solves a Problem Is The Problem Is: A)How Far Above Its Initial Position Does

The Equation That Solves a Problem Is The Problem Is:
A)How Far Above Its Initial Position Does