Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

A skateboarder shoots off a ramp with a velocity of 5.0 m/s, directed at an angle of 57° above the horizontal. The end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?

Sagot :

By using the suvat equations:
(a) [tex]v^2=u^2+2as[/tex] for the vertical components, so
 [tex]0^2=(5sin57)^2+2*-9.81*s[/tex]
therefore s=0.896m
However, since the ramp is 1.2m above the ground, the skateboarder is 0.896+1.2=2.096m above the ground.

(b) Work out the time when the skateboarder reaches the highest point
v=u+at for the vertical components, so 0=5sin57-9.81*t, therefore t=0.427s
Now using the horizontal components: s=ut, so s=5cos57*0.427=1.164m
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.