Ek Golf Club

ek golf club

Physics question 30 help please?

One man used a 7-iron to cut a ball on a golf tee 45.7g high. Initially, the ball travel horizontally on the floor and hit a ditch 120 m of sand, where he was beaten. The starting box is 6.00 m above the Sandtrap. Determine the dynamic on the ball when he was struck by the club (with formula FT = mv). The answer to this question is 4.98 N. s, but for some reason I keep getting 0.498. What I did was using the formula Ep = mgh energy, so I used 1/2mvsquared Ek = for the value of V that I found 10.84988479 m / s, which is then multiplied by the mass is 0.0459 kg. 0498 Then I had my answer. Can someone please tell me what's wrong?

You use the energy equation wrong. When the ball leaves the tee which has a height "h" v the velocity and (o). When he hits the Sandtrap with a final velocity "V" of the energy equation would be: (01/02) mv (o) ^ 2 + mgh = (1.2) mv ^ 2 and you do not have enough information to v (o) Use the fact that v (o) is horizontal for time travel in altitude "h" is h = (1 / 2) gt ^ 2 where is the time in the known distance for horizontal v (o): V (o) = x / t (since the horizontal velocity is constant) This will give you the right speed for use in the formula for momentum.

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