Asked by Zafr
An object is dropped from a height H. During the final second of its fall, it traverses a distance of 57.8 m. What was H?
Answers
Answered by
bobpursley
you have posted three of these motion problems without any demonstrated effort. We will be happy to critique your efforts.
Answered by
Damon
h positive up
h(t) = Hi + Vi t -4.9 t^2
0 = 57.8 + Vi*1-4.9*1^2
Vi = -52.9 m/s at 57.8 m
so how far did it fall to get to that speed and height?
v = -gt
-52.9 = -9.8 t
t = 5.40 seconds
how far did it fall in those first 5.4 seconds? 4.9 * 5.4^2
57.8 = Hi - 4.9 * 5.4^2
Hi = 143 + 58 = 201 m
h(t) = Hi + Vi t -4.9 t^2
0 = 57.8 + Vi*1-4.9*1^2
Vi = -52.9 m/s at 57.8 m
so how far did it fall to get to that speed and height?
v = -gt
-52.9 = -9.8 t
t = 5.40 seconds
how far did it fall in those first 5.4 seconds? 4.9 * 5.4^2
57.8 = Hi - 4.9 * 5.4^2
Hi = 143 + 58 = 201 m