Answer to Question #104974 in Mechanics | Relativity for Ceaser

Question #104974
A daredevil on a motorcycle leaves the end of a ramp with a speed of 35.0 m/s .If his speed is 33.0 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.
1
Expert's answer
2020-03-09T11:15:18-0400

Let the angle at which the motorcycle leaves the ramp makes an angle "\\theta" with the horizontal

the horizontal component= "35cos\\theta"

vertical component = "35sin\\theta"

At the maximum height the vertical component of the velocity becomes zero

"35cos\\theta=33\n\\\\cos\\theta=\\dfrac{33}{35}" "sin\\theta=\\dfrac{11.67}{35}"


vertical component ="35sin\\theta =11.67m\/s"

at maximum height vertical component of velocity is zero

So applying formula

"v^2=u^2+2gS"

"0=(11.67)^2+2(-10)S \\\\S=136\/20=6.8 m"


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Comments

Assignment Expert
14.06.21, 19:59

theta = ArcCos(33/35) = 19.5 degrees, then 35*sin(\theta) = 11.67


How sin comes as 11.67?
14.06.21, 05:16

How sin comes as 11.67?please ans fast

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