11:54 PM .oooo Fr T-Mobile T 88% Lawrence Technological University In the Applic
ID: 1535862 • Letter: 1
Question
11:54 PM .oooo Fr T-Mobile T 88% Lawrence Technological University In the Application part of this experiment, launcher ejects a ballupward as shown. You are working on the equation to solve its flight time. Based on the coordinates selection shown below, which of the folowing formulas have the correct sign for every single term? (Hint: what is the positive direction? where is the origin?) In this lab, you set up the track as shown below, and you measured Am1.35m, B 0.148m, What is the angle d in unit of degree? A ball rolls off a track at a height 1.45 meters above the ground, its exit velocity is 2.15 m/s at an angle 6.44 degrees below horizontal, how long is its fight time inExplanation / Answer
Question 5
the equation for the time of flight is
the ball is like horizontal projectile whose motion of equation is
yt = 0+ v1y*t +0.5 gt^2
if we take the origin at the point of projection then the ball motion is downward taking it as -ve
-yt = 0 -v1y*t-0.5gt^2
multiplying eith -ve sign on both sides s
yt = 0+ v1y*t +0.5 gt^2 ------------->>> answer
Question 6
Given A = 1.35 m, B = 0.148m
A is hypotenus and B is opposite side so
sin theta = oposite side / hypotenus
theta = arc sin (B/A)
theta = arc sin (0.148/1.35)
= 6.29 degrees ------------------>>> Ans
Question 7
similar to question 6
the equation is yt = 0+ v1y*t +0.5 gt^2 ------------->>> Ans
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