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problem says:

A baseball player throws a ball with a mass of $0.150kg$ with a speed of $40.0 m/s$ and an angle of $30$ degrees. What is the kinetic energy of the ball when it reaches the highest point on its trajectory?

So I'm guessing this is a projectile motion - kinetic energy - energy transfer problem. I know how to get the time in which it reaches the highest point, I also know how to get $y$ of that point, but to transfer these things into kinetic energy is really difficult for me. I'm guessing that since the change on kinetic energy is equal to the work done over the object, I must get that, and I'm thinking of an Integral for the area under the curve, but I don't know how to get the function for force, or what distance should I use, maybe time? please help

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    $\begingroup$ Rdog60 answer is correct, provided we can assume that there is no air drag (friction) involved. Since there is no reference to air drag in the problem statement, you can only assume air drag is to be neglected. $\endgroup$
    – Bob D
    Commented Sep 7, 2019 at 21:05

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For this we can use that the kinetic energy of the ball will be $\frac{1}{2}mv^2$

The velocity of the ball at its highest point will be all in the $x$ direction, as it its maximum point along its trajectory, it has no vertical velocity. Its horizontal velocity at this point is equal to its initial horizontal velocity, as no force has acted on the ball in the $x$ direction. So, $V=40\cdot\cos(30^°)$. Plug this, along with the mass of the ball, into the $\frac{1}{2}mv^2$ equation and you should be good.

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  • $\begingroup$ wow, yes that's 90 J, which is the correct answer according to the book, thx, I think I was overthinking the problem $\endgroup$
    – rorod8
    Commented Sep 7, 2019 at 20:36
  • $\begingroup$ No problem my guy! $\endgroup$
    – Rdog60
    Commented Sep 7, 2019 at 20:39
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the fast way by Rdog60 relies on the important fact that the total KE can be split into say horizontal and vertical "parts" (not components of a vector) KE tot= KEx + KEy where KEx = (1/2)m Vx^2 etc basically coming from pythagoras. KEx=(1/2)(3/20)(40 (root 3)/2)^2=(3/40)(20r3)^2= 90 Joule

Note the questioners method can also get the right answer but rather lengthier. initial vel = (Ux, Uy)=40(cos30, sin30) where cos30=(root3)/2 sin30= 1/2, Uy=20 m=3/20 kg initialKE= (1/2)mU^2=(3/40)40^2=120 J final vel at high point Vx=Ux, Vy=0 suvat for vertical y motion gives Vy = Uy -gT ... where g=10m/s/s approx T= (40/2)/10 = 2 sec final position=(x,y) x=Ux.T , y = Uy.T - (1/2)gT^2 =20(2) -5(4)= 20 work done = force . distance = Integral Fy dy = Integral mg dy = mgy=(3/20)10(20)=30 finalKE = initialKE - work lost = 120- 30 = 90 Joule

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  • $\begingroup$ You can use MathJax to typeset equations. $\endgroup$
    – PM 2Ring
    Commented Sep 8, 2019 at 7:45

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