<p>A skateboarder shoots off a ramp with a velocity of <span style=\"color: #ff0

Question

<p>A skateboarder shoots off a ramp with a velocity of <span>6.7</span> m/s, directed at an angle of <span>60</span><span>&#176;</span> above the horizontal. The end of the ramp is <span>1.1</span> m above the ground. Let the <em>x</em> axis be parallel to the ground, the +<em>y</em> direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp.</p>
<div class="indent">(a) How high above the ground is the highest point that the skateboarder reaches?<br />answer is in&#160;m<br /><br />(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?<br />answer is in m</div>

Explanation / Answer

U = 6.5m/s at 62° Ux = 6.5cos62 = 3.052m/s Uy = 6.5sin62 = 5.739m/s g = 9.8m/s² a) max height is when Vy = 0 so 0 = Vy² = Uy² - 2gy gives the max height as y = Uy²/(2g) = 5.739²/19.6 = 1.68m b) also if Vy=0 then 0 = Vy = Uy - gt so the time is takes to reach max height is t = Uy/g then the horizontal distance is x = Uxt = UxUy/g = 1.79m

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