Projectile Motion Problem: Tennis Ball Clearance Calculation

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SUMMARY

The discussion focuses on calculating the initial velocity of a tennis ball hit at an angle of 3 degrees to clear a net 0.330m high, located 12.6m away. Participants emphasize the need to separate the motion into horizontal and vertical components, using kinematic equations for each direction. The key equations discussed include the vertical motion equation \(s = v_0 t + \frac{1}{2} a t^2\) and the horizontal motion equation where acceleration is zero. The solution requires finding the initial velocity components using trigonometric functions.

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  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic knowledge of trigonometry for resolving velocity components
  • Concept of gravitational acceleration (g = 9.8 m/s²)
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  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to resolve vectors into components using sine and cosine functions
  • Explore the concept of maximum height in projectile motion
  • Practice solving similar problems involving angles and distances in projectile motion
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m_scott
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Homework Statement


A tennis player standing 12.6m from the net hits the ball at 3 degrees above the horizontal. To clear the net, the ball must rise at least .330m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racquet?


Homework Equations


I tried using v2=v02+2a(x-x0)


The Attempt at a Solution


I have no clue
 
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welcome to pf!

hi m_scott! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)

you need two equations, one for the x and one for the y direction …

then eliminate t …

what do you get? :smile:
 
I still don't completely understand. What formulas am i supposed to use? And I am not sure what to do with the .330m

I set it up as: 0(final velocity)= v02+2(-9.8)(12.6) and got 15.7 but i know that's wrong...
 
Hi m_scott.
You must see the motion of the ball as a composition of two kinds of motion: one in the horinzontal axis and another in the vertical axis.
Which kind of motion on the x axis?
Which kind of motion on the y axis?
Try to write the equations of motion for each axis.
And from it, up to you...
 
uhh? :redface: forget the final velocity …

nobody asked for it! :wink:
 
what equation should i use to find motion for the x and y axis?
 
What kind of motion do you think that has the ball on the y direction ?
 
well, you're only given s and vi and a (a = 0 for the x direction, = -g for the y direction), and you want to find t for both directions and then eliminate it …

so which of the standard constant acceleration equations do you think you should use? :smile:
 
s=v0t+(1/2)at2??

And if so, do i plug in -9.8 for a and 0 for v0?

i got 1.6s Is that for the x or y axis?
 
  • #10
m_scott said:
s=v0t+(1/2)at2??

yes :smile:
And if so, do i plug in … 0 for v0?

no, v0 will be the initial component of velocity in the y direction
 
  • #11
Ok. How do I find the initial component of velocity in the y direction? Btw, sorry for all the newb questions. This is my 3rd day in physics and I am kinda lost...
 
  • #12
total velocity times sin3°

(and the initial x component is total velocity times cos3°)

haven't you done components? :confused:
 
  • #13
But the problem doesn't give me the total velocity
 
  • #14
that's right, it asks you for the total velocity …

so call that v, and solve :wink:
 

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