Solving a Golf Ball Problem: V=105/3.8 & Y = 70.76 sin 27.63

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The discussion revolves around calculating the initial velocity of a golf ball struck with a five iron, which lands 105 meters away in 3.8 seconds. The initial calculation provided was V=105/3.8, resulting in a horizontal speed of 27.63 m/s. A participant pointed out that this value represents only the horizontal component and suggested using the vertical motion formula to find the vertical component of the velocity. By applying the correct formulas and values, the original poster was able to resolve the problem successfully. The conversation highlights the importance of considering both horizontal and vertical components in projectile motion calculations.
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Well I've been working on these last two problems and I just can't figure them out. Can someone just point me in the correct direction?

1.) A golf ball is struck with a five iron on level ground. It lands 105.0m away 3.80 s later. What is the magnitude and the direction of the initial velocity?(Neglect air resistance.)

This is what I have for this problem: V=105/3.8 = 27.63m/s
Then I used y = Vyot - (1/2)g t^2 the answer from this was 70.76 for which I then used 70.76 sin 27.63 to get an angle.

thanks in advance for the help

kwesi
 
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Check your working again for the velocity of 70.67.
 
gen1pro said:
V=105/3.8 = 27.63m/s

Careful there, that's just the (constant) speed in the horizontal direction. If you want to use the formula y(t) = V_{y}(0)\cdot t - \frac{1}{2}gt^{2}, try to fill in the values for the moment the ball touches down again. You should end up with V_{y}(0). Then you'll have both components of the initial velocity vector, and from there it's just a small step to complete the answer.
 
thanks for the info I was able to figure it out with your help.
 
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