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Projectile Problem

  • Thread starter Lord Dark
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  • #1
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Homework Statement


hi guys ,, got this problem about projectile ,,
A ball is thrown from the top of a building with a velocity of 40 m/s at an angle of 53
with the horizontal. After 2 s, it is seen to be at height of 84 m above the ground.
a) Find the height of the building
b) At which other time will the ball again be at a height of 84 m?
c) If the ball hits a wall at a height of 52 m above the ground, what is the distance of the
wall from the building?
d) Find the magnitude and direction of the velocity of the ball when it reaches the wall

Homework Equations





The Attempt at a Solution


i tried to get the height of the building by 84+h(b)=40sin(53)*2-9.8*4 ,, and I got h(b)=59.31m ,, anyway ,, the way is right ?? ,, and about (b) i tried to get the second time ,, but i get it negative (59.31+84=40sin(53)x-9.8*x^2 = 9.8x^2-40sin(53)x-143.31 and i get x=5.7 , x=-2.5 and i think it's wrong because none of them equal 2 ,, and one of them should equal 2 ,, any ideas what did I do wrong here ?
 

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Answers and Replies

  • #2
Doc Al
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i tried to get the height of the building by 84+h(b)=40sin(53)*2-9.8*4 ,, and I got h(b)=59.31m ,, anyway ,, the way is right ??
Your expression for vertical position is not quite right. What's the general formula?
 
  • #3
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delta(y)=Voyt-0.5*g*t^2 ,, (I forgot half) thanks :D
 
  • #4
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now i get the height 72m ( which i think it's too high ) and i still cant get the second time,
 
  • #5
Doc Al
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44,877
1,129
now i get the height 72m ( which i think it's too high ) and i still cant get the second time,
Show the final equation that you solved, with the numbers that you used.
 
  • #6
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123.71=40sin(53)t-0.5*9.8*t^2 ,, now i don't even get an answer (they are imaginary numbers)
 
  • #7
Doc Al
Mentor
44,877
1,129
123.71=40sin(53)t-0.5*9.8*t^2 ,, now i don't even get an answer (they are imaginary numbers)
Show me the equation that you used to solve for the height of the building.
 
  • #8
121
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123.71=40sin(53)t-0.5*9.8*t^2 ,, now i don't even get an answer (they are imaginary numbers)
 
  • #9
Doc Al
Mentor
44,877
1,129
123.71=40sin(53)t-0.5*9.8*t^2 ,, now i don't even get an answer (they are imaginary numbers)
Show me the equation that you used to solve for the height of the building. (There should be some variable representing the height in there somewhere.)
 
  • #10
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84+h(b)=40sin(53)*2-0.5*9.8*4 (Y-Yo=Voy*t-0.5*g*t^2)
 
  • #11
Doc Al
Mentor
44,877
1,129
84+h(b)=40sin(53)*2-0.5*9.8*4 (Y-Yo=Voy*t-0.5*g*t^2)
Compare the first equation to the one in parentheses. (Careful with signs.)
 
  • #12
121
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....... Did it ,, (84-39.71)=40sin(53)t-0.5*9.8*t^2 and i get t1=1.9 = 2s t2=4.5 s ,, about (c) = first I got t=6.1s then (X=Vox*t) =147.1m
(d) Vx=24 m/s (never change) Vy=-27.8 then get lVl and get the angle which is -49

thanks :D
 

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