Suppose the departure angle in the figure below is 41.4 Dashed arrows show how f

Question

Suppose the departure angle in the figure below is 41.4

Dashed arrows show how far the dart and monkey have fallen at specific times relative to where they would be without gravity. At any time, they have fallen by the same amount.. Without gravity The monkey remains in its initial position. The dart travels straight to the monkey. Therefore, the dart hits the monkey. The monkey falls straight down. At any time I. the dart has fallen by the same amount as the monkey relative to where either would be in the absence of gravity: Delta y dart = Delta y monkey = -1/2 gt2 Therefore, the dart always hits the monkey.

Explanation / Answer

Suppose the departure angle is 41.4 degrees and the distance d is 3.23 m.
Where will the dart and the monkey meet if the initial speed of the dart is:

a) 14.5 m/s

Solve with the numbers given and, by default, you reach the conclusion that the two either hit or don't based on what information they give you.

So here using V0x = V0cos(theta) and V0y = V0sin(theta) I get:

V0x = 10.87 m/s
V0y = 9.589 m/s

since d = 3.23 m

x = x0 +v0xt

3.23m = 10.87 m/s*t

t = 0.2971 s.

Then where will the two meet?

y = y0 +V0yt -4.9t^2

Plug the appropriate numbers in and solve for "y". (y0 is = 0)

y = 2.41m

b) 8.5 m/s

So here using V0x = V0cos(theta) and V0y = V0sin(theta) I get:

V0x = 6.37 m/s
V0y = 5.62 m/s

since d = 3.23 m

x = x0 +v0xt

3.23m = 6.37 m/s*t

t = 0.507 s.

Then where will the two meet?

y = y0 +V0yt -4.9t^2

Plug the appropriate numbers in and solve for "y". (y0 is = 0)

y = 1.589 m

c) 4.0 m/s

So here using V0x = V0cos(theta) and V0y = V0sin(theta) I get:

V0x = 3 m/s
V0y = 2.64 m/s

since d = 3.23 m

x = x0 +v0xt

3.23m = 3 m/s*t

t = 1.07s.

Then where will the two meet?

y = y0 +V0yt -4.9t^2

Plug the appropriate numbers in and solve for "y". (y0 is = 0)

y = -2.785 m

Negative sign indicates that they never meet.

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