You are swinging on the swing set at the park, and are trying to leap off of the

Question

You are swinging on the swing set at the park, and are trying to leap off of the wing so that you land in the grassy part of the schoolyard beyond the sand.

a. Suppose you leave the swing 3.5m above the ground at an angle of 40 degrees to the horizontal, with a speed of 5.0 m/s. What is your velocity in the x-direction just before you land?

b. What will be your maximum height above where you leave the swing?

c. You are hoping to land your jump 10 m from where you launch off of the swing. Given the initial conditions, how far will you go and wil you make your initial goal?

Explanation / Answer

a. initial horizontal velocity=5*cos(40)=3.83 m/s

as there is no acceleration in horizontal direction, veloicty in x direction will be same as initial horizontal velocity

hence answer is 3.83 m/s.

b.

initial vertical velocity=5*sin(40)=3.214 m/s

vertical acceleration=-9.8 m/s^2 (as it is in downward direction)

at maximum height, vertical veloicty=0

using the formula:

final vertical veloicty^2-initial vertical velocity^2=2*vertical acceleration*height

==>0^2-3.214^2=-2*9.8*height

==>height=3.214^2/19.6=0.527 m

hence maximum height above the point where you leave the swing is 0.527 m

c.

displacement in vertical direction for reaching ground=-3.5 m

using the formula:

displacement=initial vertical veloicty*time+0.5*vertical acceleration*time^2

==>-3.5=3.214*t-0.5*9.8*t^2

==>t=1.2345 seconds

then distance covered along horizontal distance=horizontal velocity*time taken to reach ground

=3.83*1.2345=4.7284 m

hence you wont be able to make your initial goal.

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