I just want to make sure I did this right. #K24 A freely-falling object falls in
ID: 1653829 • Letter: I
Question
I just want to make sure I did this right.
#K24 A freely-falling object falls in a laboratory near the surface of the Earth. It starts from rest at y = 0.00 m . Follow the instructions below. (a) Calculate the velocity of the particle at 1 = 0.950 s . (b) Calculate the position of the particle at 1-0950 Freely-falling motion is an important example of motion with constant acceleration, and the acceleration is known: a,-g. Therefore, the constant acceleration equations may be used, adapted for vertical motion (+y up). Here is a way to organize the information and solve the problem systematically. It is how problems of this type will be solved in PHYS 120. That is, please make these lists. (a) KNOW: yo = 0.00 m, voy-0.00 m/s, ay-g-9.80 m/s*, 1-0.950 s WANT: v, USE: vy aThis equation involves the known and wanted quantities. It does not contain more than one unknown quantity. This constant acceleration equation is already solved symbolically for the requested quantity (v.). Substitute the numbers and compute the value. (b) KNOW: Y) = 0.00 m. voy0.00 m/s, a, =-g =-9.80 m/s, t = 0.950 s WANT: y USE: y = y0 + v0/t-a,T This equation involves the known and wanted quantities. It does not contain more than one unknown quantity. Once again, this constant acceleration equation is already solved symbolically for the requested quantity (y). Substitute the numbers and compute the value. Note that if the result of (a) is used, then other equations could be used.Explanation / Answer
a] Since upward motion is chosen as the position direction therefore,
v = u + at = u - gt
so, after t = 0.950s, v = 0 - 9.8(0.95) = - 9.31 m/s
this is the velocity after 0.950 seconds.
b] Use y = yo + ut + (1/2)at2 = 0 - (1/2)gt2
so, after t = 0.95s; y = 0 + 0 - (1/2)(9.8)(0.95)2 = - 4.42225 m.