Re-open discussion: Projectile motions help
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for the original question:
How do I answer this question?
a) A rugby player is aiming for a conversion. He kicks the ball at 15ms−1 at an angle of 50∘ to the horizontal. At the time, he is 20m from the posts. How much time will the ball take to reach the posts?
b) How high will the ball be when it reaches the posts?
I've been struggling on this question for ages. So any answers and help would be really helpful...
the answer was explained this : (link: https://www.thestudentroom.co.uk/showthread.php?t=5272002&p=76809304)
Okay, so calculating the horizontal component of V gives '9.6418.. m/s' which then gives a a time of 2.074 s from '20/9.6418'
Then moving onto part b, I calculated the the vertical component of V which is '11.49 m/s'.
Subbing all this into the suvat equation of 's = ut + at^2 and using the value of time from the earlier part along with '11.49 m/s' gives an answer of 2.8 to 1dp.
why s = ut + at^2 used, why not s = ut + (1/2)at^2? pls explain.
for the original question:
How do I answer this question?
a) A rugby player is aiming for a conversion. He kicks the ball at 15ms−1 at an angle of 50∘ to the horizontal. At the time, he is 20m from the posts. How much time will the ball take to reach the posts?
b) How high will the ball be when it reaches the posts?
I've been struggling on this question for ages. So any answers and help would be really helpful...
the answer was explained this : (link: https://www.thestudentroom.co.uk/showthread.php?t=5272002&p=76809304)
Okay, so calculating the horizontal component of V gives '9.6418.. m/s' which then gives a a time of 2.074 s from '20/9.6418'
Then moving onto part b, I calculated the the vertical component of V which is '11.49 m/s'.
Subbing all this into the suvat equation of 's = ut + at^2 and using the value of time from the earlier part along with '11.49 m/s' gives an answer of 2.8 to 1dp.
why s = ut + at^2 used, why not s = ut + (1/2)at^2? pls explain.
Reply 1
I'm going to be honest, I don't really understand your question, so I'm going to go through both parts with working and then you tell me what you're confused about.
a) The horizontal component of the velocity will be 15cos(50). Plugging this into s = vt gives 20 = 15cos(50)t so t = 2.074298436 as above.
b) At the start, the vertical component of the velocity will be 15sin(50). Acceleration will be 9.81ms^-2 downwards. Using s = ut + 1/2at^2 gives s = 15sin(50)*2.07... - 4.905(2.07...^2) and s = 2.73, 2.7 to 1 d.p. (I think original responder had rounding errors due to not leaving u as 15sin(50)). s = ut + 1/2at^2 was used, I think they just must've made a mistake writing it.
a) The horizontal component of the velocity will be 15cos(50). Plugging this into s = vt gives 20 = 15cos(50)t so t = 2.074298436 as above.
b) At the start, the vertical component of the velocity will be 15sin(50). Acceleration will be 9.81ms^-2 downwards. Using s = ut + 1/2at^2 gives s = 15sin(50)*2.07... - 4.905(2.07...^2) and s = 2.73, 2.7 to 1 d.p. (I think original responder had rounding errors due to not leaving u as 15sin(50)). s = ut + 1/2at^2 was used, I think they just must've made a mistake writing it.
Reply 2
kanji13
OP
4
ok, thank you, I got it now. The original responder must've made a mistake when writing the equation.
And, u use -9.8 for "a" cuz it's downwards. Ty 👌🙏
basically, the method is:
to find the height (how high the ball will be at the posts) when u have a q that tells smth thrown/kicked/fired at an angle to horizontal.
And, u use -9.8 for "a" cuz it's downwards. Ty 👌🙏
basically, the method is:
1.
s = ut + (1/2)at2
2.
u = vertical component
3.
a = -9.8 (cuz downward)
4.
time from what we calculated before
Reply 3
kanji13
OP
4
Original post
by kanji13ok, thank you, I got it now. The original responder must've made a mistake when writing the equation.
And, u use -9.8 for "a" cuz it's downwards. Ty 👌🙏
basically, the method is:
to find the height (how high the ball will be at the posts) when u have a q that tells smth thrown/kicked/fired at an angle to horizontal.
And, u use -9.8 for "a" cuz it's downwards. Ty 👌🙏
basically, the method is:
1.
s = ut + (1/2)at2
2.
u = vertical component
3.
a = -9.8 (cuz downward)
4.
time from what we calculated before
ok, wait a min, did u get -4.9 cuz u used s = vt - (1/2)at2 ? (-1/2 x 9.8 = -4.9)
Reply 4
Original post
by kanji13ok, wait a min, did u get -4.9 cuz u used s = vt - (1/2)at2 ? (-1/2 x 9.8 = -4.9)
I used a = -9.81, the 4.905 is there because it's after halfing if that's what you're asking. Lmk if u have any more questions (like why it's sin(50) and not cos(50) or anything else), this is a critical time in your education and understanding now will massively help later.
Reply 5
kanji13
OP
4
Original post
by AltAccount00I used a = -9.81, the 4.905 is there because it's after halfing if that's what you're asking. Lmk if u have any more questions (like why it's sin(50) and not cos(50) or anything else), this is a critical time in your education and understanding now will massively help later.
oh ok, gotchu😎👌
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