Physics problem answer check

[jai6638]

Hey there..

I was given a question in a quiz for my Physics 1 class and I solved it. However, it was marked wrong. Could anyone please highlight why it is wrong?

Q) At a hockey game, you throw your hat into the air when our team scores a point. If you throw it upwards at an angle of 35 degrees with respect to the horizontal and speed of 11 m/s:

D) Assuming you are on level ground, how far away from you does it land ?

I found the time it takes to get to the highest point to be ( .6438 s and its maximum height to be .32 m ) in the earlier questions that I have not typed in this post.

To find this answer, I used the Range equation, R=((V0)^(2) * sin(2theta))/(g)
R= (11)^(2) * sin ( 2*35) / ( 9.8 )
R = 11.6 m.

Apparently, the answer is 5.8m but I'm not sure why it should be half of my answer. The range gives you the distance it covers in the X dimension so why is it half?!

Thanks

 

[JohnCU]

speed in the x direction is 11 * cos(35)

then that number * .6438s = 5.8 m

it's a lot easier to do it that way than using the plug and chug. that is, assuming my method is correct . i just remembered that the two directions/speeds(?) are independent of another one and the 11 m/s at an angle of 35 degrees can be broken down into two different vectors and wah lah.

 

[darkxshade]

The correct answer is: far enough for it to land on the ice.

 

[Beattie]

Originally posted by: JohnCU
speed in the x direction is 11 * cos(35)

then that number * .6438s = 5.8 m

it's a lot easier to do it that way than using the plug and chug. that is, assuming my method is correct . i just remembered that the two directions/speeds(?) are independent of another one and the 11 m/s at an angle of 35 degrees can be broken down into two different vectors and wah lah.

You only calculated the distance from you the hat is at when it reaches the apex of the curve.

 

[JohnCU]

hmm, i thought the time he gave was the time in flight, nevermind. multiply my answer by 2 and your teacher is wrong.

 

[Beattie]

Originally posted by: jai6638
Hey there..

I was given a question in a quiz for my Physics 1 class and I solved it. However, it was marked wrong. Could anyone please highlight why it is wrong?

Q) At a hockey game, you throw your hat into the air when our team scores a point. If you throw it upwards at an angle of 35 degrees with respect to the horizontal and speed of 11 m/s:

D) Assuming you are on level ground, how far away from you does it land ?

I found the time it takes to get to the highest point to be ( .6438 s and its maximum height to be .32 m ) in the earlier questions that I have not typed in this post.

To find this answer, I used the Range equation, R=((V0)^(2) * sin(2theta))/(g)
R= (11)^(2) * sin ( 2*35) / ( 9.8 )
R = 11.6 m.

Apparently, the answer is 5.8m but I'm not sure why it should be half of my answer. The range gives you the distance it covers in the X dimension so why is it half?!

Thanks

Your answer is right... Ask the teacher about that one.

Edit = actually, it's slightly farther than that because you probably have an initial height that your hand threw the hat from whereas it is landing on the ground.

 

[jdobratz]

Breaking down the component vectors of the velocity I get 6.31 m/s in the vertical and 9.01 m/s on the horizontal. I think you got the time correct, I got the same value of 0.64s. So, using this equation:

x = vix*t + 1/2*ax*t^2

where:

x = horizontal displacement
vix = inital vertical velocity
ax = horizontal acceleration
t = time

I get:

x = 9.01*0.64 + 1/2*0*0.64^2
x = 5.7664 + 0
x = 5.7664 or about 5.8m

Im quite rusty on physics and was using googled equations, so I could be completely wrong.

After further review, I messed the time up, it should be twice that I think. So your answer appears correct.

 

[jai6638]

Thanks for the responses guys. Will talk to the TA when I see him next week.

Breaking down the component vectors of the velocity I get 6.31 m/s in the vertical and 9.01 m/s on the horizontal. I think you got the time correct, I got the same value of 0.64s. So, using this equation:

x = vix*t + 1/2*ax*t^2

where:

x = horizontal displacement
vix = inital vertical velocity
ax = horizontal acceleration
t = time

I get:

x = 9.01*0.64 + 1/2*0*0.64^2
x = 5.7664 + 0
x = 5.7664 or about 5.8m

Im quite rusty on physics and was using googled equations, so I could be completely wrong.

0.64secs is the time it takes to reach the highest point. Hence the 5.8 would have to be doubled to get the total distance.