Two marbles are launched at t =0 in the experiment illustrated in the figure bel
ID: 1533333 • Letter: T
Question
Two marbles are launched at t=0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.08 m/s from a height h=0.910m. Marble 2 is launched from ground level with a speed of 5.77 m/s at an angle =45.0 above the horizontal
Part A
Where would the marbles collide in the absence of gravity? Give the x coordinate of the collision point.
Part B
Where would the marbles collide in the absence of gravity? Give the y coordinate of the collision point.
Part C
Where do the marbles collide given that gravity produces a downward acceleration of g=9.81m/s2? Give the x coordinate.
Part D
Where do the marbles collide given that gravity produces a downward acceleration of g=9.81m/s2? Give the y coordinate.
Express your answers to three significant figures and include appropriate units.
5607638 zontally with a speed of 4.08 m/s from a height h 0.910 m. Marble 2 is launched from ground level with a sExplanation / Answer
in the absence of gravity the marble 1 travels horizontally
the time taken by the marble2,to reach the height is t
h = u2*t
0.91 = (5.77*sin(45)*t)
t = 0.22 sec
so after 0.22 sec the two marbles will collide
A) hence the required x -coordinate is 4.08*0.22 = 0.8976 m
B) y-coordinate is y = 0.91 m
C) if we consider the gravity,let after t sec the collision takes place
horizontal distance travelled by marble 1 in t sec = horizontal distance travelled by marble 2 in t sec
U1*t = u2*cos(45)*t
vertical distance travelled by marble 1 in t sec + vertical distance travelled by marble 2 in t sec = 0.91
0.5*g*t^2 + u2*sin(45)*t - (0.5*g*t^2) = 0.91
0.91 = 5.77*sin(45)*t
0.91 = 5.77*sin(45)*t
t = 0.223 sec
so x-coordinate is x = u1*t = 4.08*0.223 = 0.909 m
D) y-coordinate is y = u2*sin(45)*t - (0.5*g*t^2)
y = (5.77*sin(45)*0.223) - (0.5*9.8*0.223^2) = 0.666 m
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