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Test Questions

  1. Jul 3, 2005 #1
    I have some test questions that I got wrong that I need to know how to solve for the final. Any help would be appreciated.

    1)A place kicker must kick a football from a point 33.5 m from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 19.5 m/s at an angle of 38.0° to the horizontal.
    a)By how much does the ball clear or fall short of clearing the crossbar? (Enter a positive value for clearance and a negative value for falling short.)

    Do I use the equation Ymax=Voy^2/2a ? Because the correct answer should be -0.19 m

    b)What is the y-component of the ball's velocity as it reaches the crossbar?

    I figure I use Vf = Vi +a(2*distance)/(Vf +Vi) but I don’t know how to get distance from the first part to do this. The correct answer should be -9.38 m/s

    2) A ball is projected horizontally from the edge of a table that is 1.08 m high, and it strikes the floor at a point 1.24 m from the base of the table.
    a)What is the initial speed of the ball? 2.64 m/s

    b)How high is the ball above the floor when its velocity vector makes a 47.6° angle with the horizontal?
    6.53×10-1 m

    I got part a) right by using t=sqRoot(2h/g) and d/t=v but I have no idea what to do with part b!

    3) One of the fastest recorded pitches in major-league baseball, thrown by Nolan Ryan in 1974, was clocked at 100.8 mi/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached a horizontal distance of 65.0 ft? (Neglect air resistance.)
    9.49×10-1 m

    I found time using the Kinematics equations then plugged it into y=.5at^2 but that doesn’t seem to work.
  2. jcsd
  3. Jul 3, 2005 #2


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    For most of these questions, you have to seperate the components of the displacement, velocity and acceleration in the x and y directions (ie horizontal and vertical) seperatley. Now it is important to remember that the acceleration in the y-direction will affect the velocity in the y-direction only.
    Let's look at question 1 first.
    The speed of the ball is given. What are it's components in the x&y directions? Can you find that out? Let them be [tex] v_x [/tex] and [tex] v_y [/tex] respectivley.

    Then, look at motion in the x-direction. What is the acceleration in the x-direction? It is zero.
    So the velocity in the x-direction is constant (is that clear?). Therefore the displacement of the ball as a function of time in the x-direction is given as
    [tex] x=(v_x)(t) [/tex]

    Now, look at the motion of the ball in the y-direction. The acceleration in the y-direction is g m/sec^2 downwards.
    So the displacement of the ball as a function of time in the y-direction is given as
    [tex] y=(v_y)(t) - 0.5gt^2 [/tex].
    Also, when the ball falls on the ground again the y-displacement is zero. Put that in the above equation to get the time of flight.
    Now the question asks for the horizontal position of the ball as the ball touches the ground. But you already know the horizontal displacement as a function of time! So plug in the value of time and you will know where the ball is when the ball hits the ground again. Is that clear?
    Let's see you do the other questions now!
  4. Jul 3, 2005 #3


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    I'd start with the following equations

    a) h = h_0 + v_0y*t + 1/2at^2
    b) v_y = v_0y + 1/2at^2

    a) h = h_0 + v_0*t + 1/2at^2
    b) [tex]tan\alpha = \frac{v_y}{v_x}[/tex]

    3) h = h_0 + v_0y*t + 1/2at^2
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