Solved

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

Question 33

Multiple Choice

The equation that solves a problem is 6.4 m=20 m+3.0ms(2.0 s) 4.9ms2(2.0 s) 26.4 \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?

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions