(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 52.4 g ball is thrown from the ground into the air with an initial speed of 16.3 m/s at an angle 27.4^{o}above the horizontal

a) What are the values of kinetic energy of the ball initially and just before it hits the ground

b) Find the corresponding values of the momentum (magnitude and direction) and the change in momentum.

c) Show that the change in momentum is equal to the weight of the ball multiplied by the time of flight, and thereby find the time of flight.

2. Relevant equations

KE = mv^{2}/2

p = mv

J = p_{2}- p_{1}= F[tex]\Delta[/tex]t = m*a*t

3. The attempt at a solution

a) 0.0524*16.3^{2}/2 = 6.96 J both going up and coming down

b) 0.0524*16.3 = 0.854 So p_{1}0.854 @ 27.4^{o}

p_{2}0.854 @ -27.4^{o}

[tex]\Delta[/tex]p = p_{2}- p_{1}= 1.71

c) Please check a and b - because I just can't seem to get this to work

I've used h_{max}= V^{2}SIN^{2}[tex]\theta[/tex]/2g

and t = squareroot (2*h_{max}/g) to determine that t should equal 0.76s

but every time I try to use my change of momentum (b) to get this answer they don't agree.

Thanks in advance

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# Momentum/impulse check & help needed

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