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 swing so that you land in the grassy part of the school yard beyond the sand. A. Suppose you leave the swing 3.5 m 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 will you make your initial goal?

Explanation / Answer

x velocity will not be altered so

velocity = 5cos(40) =3.83022221559 m/s

B)

0=5sin(40) - 9.81 t

t=5sin(40)/9.81

= 0.32761855743 sec

h= 5sin(40)*t -0.5* 9.81 *t^2

=5sin(40)*0.32761855743 -0.5*9.81*0.32761855743^2

=0.52647287355 meter from initial point

after jump has come to same level now velocity is downwards

3.5 = 5sin(40) t + 0.5 *9.81*t^2

t=0.5784120786188789 s

total time of journey to fall to ground = 0.5784120786188789+2*0.32761855743

1.23364919348 sec

so distance travelled in x =1.23364919348 *5cos(40) =4.72515054712 m

Goal is not reached

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