Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
First, find the time that it takes for the sparrow to travel 0.500m at a speed of 1.80 m/s. Since speed is the ratio between distance and the time taken to travel that distance, then:
[tex]v=\frac{d}{t}[/tex]Isolate t from the equation:
[tex]\Rightarrow t=\frac{d}{v}[/tex]Replace d=0.500m and v=1.80 m/s to find the time:
[tex]t=\frac{0.500m}{1.80\frac{m}{s}}=0.2777\ldots s[/tex]During that time, the sparrow falls vertically according to the equation:
[tex]h=\frac{1}{2}gt^2[/tex]Where g is the gravitational acceleration on Earth:
[tex]g=9.81\frac{m}{s^2}[/tex]Replace the value of g and t=0.2777...s to find how far does the sparrow fall:
[tex]h=\frac{1}{2}(9.81\frac{m}{s^2})(0.2777\ldots s)^2=0.378472\ldots m[/tex]If the initial speed is increased, it will take less time to travel 0.500m, and then, the falling time will be decreased, and so the distance of fall would decrease. However, the horizontal speed does not affect the distance of fall if the fall time is the same (but the horizontal displacement would increase).
Therefore, the answers are:
a) The sparrow falls a distance of 0.378m
b) The distance of fall would decrease since it would take less time to travel 0.500m.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.
