Fence Height for Detached Race Car Wheel at 45m/s

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a wheel becomes detached from a race car, the wheel is traveling at 45m/s at an angle of 35 deg. a fence is erected 25m from the track. ignoring air resistance how high should the fence be?



gravity 9.8



ive tried drawing a diagram, with angle of 35 deg. i believe it is a trig question
 
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What will be the vertical and horizontal components of velocity using trig?
 
see that's where I am stuck i need to know how to calculate this:
if i know the angle and the adjacent,
35 deg, 25m and i want to the height, opposite, 35 x 25 as in Tan x adj
 
Ahh I see. You misunderstand slightly, its not assimple as that. The wheel does not travel in a straight line since it is subject to gravity and that is why you need to use kinematic equations. You use trig to find the components of the velocity vector (i.e. the magnitude of the velocity is the hypotenuse of a right angled triangle). The components tell you the horizontal and vertical velocity of the wheel to which you can apply the kinematic equations.
 
i was lead to believe it would travel in a straight line as its traveling at 45m/s and the fence is only 25m away
 
chrisridgwell said:
i was lead to believe it would travel in a straight line as its traveling at 45m/s and the fence is only 25m away

By who? If that is the case why were you given the gravitational acceleration of g?
 
sorry if i have confused it, this is an assignment question which i got late as i missed a lesson
the question says that the wheel is traveling at 45m/s at 35 deg. the fence is 25m away, how high does it have to be to prevent it hitting the crowd. then in brackets it just says take gravitational field strength to be 9.8 N/kg
 
The you will definitely need to take gravity into account. So begin by finding the components of the wheels velocity and then use the appropriate kinematic equations to see how high the wheel will be when it reaches the fence.
 
what do you mean by components? I am sorry I am new to physics
 
Like I said previously consider the vector as the hypotenuse of a right angled triangle. The other two sides of the triangle will then be the components of the vector which you can work out using trigonometry. Read the following page for more information on vectors in general including resolving components.

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
 
im stuck now, as the horizontal is flat so it is at 0 degress, and i can't calculate the other side as i don't have the information to.
 
Both sides can be calculated from the hypotenuse and the angle. Did you read the link I gave you about resolving components? Its just the same as calculating the length of sides of a right-angled triangle in trigonometry class.