Two marbles are launched at t =0 in the experiment illustrated in the figure bel

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 4.04 m/s from a height h=0.980m. Marble 2 is launched from ground level with a speed of 5.71 m/s at an angle =45.0 above the horizontal.

A.) Where would the marbles collide in the absence of gravity? Give the x coordinate of the collision point.

B.) Where would the marbles collide in the absence of gravity? Give the y coordinate of the collision point.

C.) Where do the marbles collide given that gravity produces a downward acceleration of g=9.81m/s2? Give the x coordinate.

D.) Where do the marbles collide given that gravity produces a downward acceleration of g=9.81m/s2? Give the y coordinate.

If you can solve all parts of the problem with the answer worked out, that would be appreciative!

Explanation / Answer

Given,

Marble 1: Velocity, v = 4.04 m/s, Height, h = 0.980 m

Marble 2: Velocity, v = 5.71 m/s, angle = 45°

In absence of gravity the marble 1 travels horizontally, if time taken by the marble 2 to reach the height is t

Then, h = v x t

0.98 = (5.71 sin(45) x t)

t = 0.243 sec

so after 0.243 sec the two marbles will collide

A) So, x -coordinate is 4.04 x 0.243 = 0.982 m

B) y-coordinate is y = 0.980 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 x t = u2 x cos(45) x t

vertical distance travelled by marble 1 in t sec + vertical distance travelled by marble 2 in t sec = 0.98

0.5 x g x t2 + u2 x sin(45) x t - (0.5 x g x t2) = 0.980

0.98 = 5.71 x sin(45) x t

t = 0.243 sec

so x-coordinate is x = u1 x t = 4.04 x 0.243 = 0.982 m

D) y-coordinate is y = u2 x sin(45) x t - (0.5 x g x t2)

y = (5.71 x sin(45) x 0.243) - (0.5 x 9.8 x 0.2432) = 0.692 m

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