I have droped an object from 20.64 ft and the time it takes to hit the ground wa
ID: 2304017 • Letter: I
Question
I have droped an object from 20.64 ft and the time it takes to hit the ground was 5.39 s. The object mass is 203 g.
Please use theses formuls to find the velocity, acceleration and position at 0.01 s and 0.02 s.
+ X Medical supply Drop Ec . ions.pdf?| ? ? ? https //ecampuswu.edu/bbcswebdav/pid: 482541 0-dt-content-rid-39743 146-1/courses/star50656.201805/Medical%205 Time Your time column in excel should show: 0.00,0.01, 0.02... until it hits the ground. Set this up in a table as the first column like figure 1. Time Position Velocity Accleration 20 32.2 Figure 1. This is an example of how the first 3 rows of your excel sheet may look Position You vill only use the initial conditions to find the first position at time 0.01 using equation 1 After that (for positions at time 0.02 and beyond), you will need to rely on the previous position, velocity and acceleration to do the rest of the calculations. Equation1 Where yo is the initial height in teet Where v is the initial velocity in ft/s Where t is the change in time in seconds (aways 0.01 seconds) where ? is the acceleration in ft/s Velocity You vill only use the initial conditions to find the first velocity value(at time 0.01) using equation 2. After that [for velocities at time 0.02 and beyond), you will need to rely on the previous velocity and acceleration to do the rest of the calculations. Equation 2 Where v, is the initial velocity in ft/s ^ 6/14/2018 ?Explanation / Answer
Initial height = 20.64 ft = 6.291 m
Assuming no air resistance, acceleration of the object = g = 9.8 m/s2
then, at t = 0.01s from the time of release, the position will be:
h = ut + (1/2)gt2 = 0 + (1/2)(9.8)(0.01)2 = 0.49 mm
the velocity is: v = u + gt = 9.8(0.01) = 0.098 m/s
at t = 0.02s, h = 0 + (1/2)(9.8)(0.01)2 = 1.96 mm
and velocity is: v = 0 + 9.8(0.02) = 0.196 m/s.