Projectile related question

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Question 1:
An aeroplane is flying at a constant horizontal velocity of 50 m s-1 at a height of 1000m. What is its horizontal distance from a target on the ground, so that a parcel released from the plane will hit the target.


I've tried using the vertical component of motion equation. But the answer is not the same. Help pls.

Question 2:
When an aircraft is 4000m above the ground and flying upwards with a velocity of 50 m s-1 at an angle of 30 degree to the horizontal. A bomb is released. Neglecting air resistance, calculate the time taken by the bomb to reach the ground

I've also tried the vertical component of motion equation. But its wrong too. Help!

Thanks~
 

Answers and Replies

  • #2
tiny-tim
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Hi crays ! :smile:

Show us your working for Question 1, so that we can see where you went wrong, and help! :smile:
 
  • #3
Kurdt
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If you show what you've done we can better identify where you need help. Essentially what you need to do for this problem is find out how long it takes to hit the ground from the height its released using the vertical component and then find out how far it will travel horizontally in that time.
 
  • #4
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Thanks for the reply guys. Here is my working so far for question 1. I've substituted the value into the vertical component. 1000m = 50t + 1/2(10t^2) and finally got a quadratic equation. And i found out t=-20 or t=10. Therefore t=10 is accepted right?

Then i proceed to applying the horizontal component of motion. x = (u cos Q)t

i know the initial velocity, u = 50 m s-1
and the t=10

but cosQ should be 90 right? Since its dropping vertically down. I'm stucked here.
 
  • #5
no initial velocity in vertical axis is 0...
the only motion initially is 50m/s in horizontal axis
 
  • #6
Kurdt
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For the horizontal component don't worry about the angle. Just work out how far it will travel at that speed in that time.
 
  • #7
tiny-tim
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1000m = 50t + 1/2(10t^2)
No, the initial vertical component is not 50, is it? :smie:
 
  • #8
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Okay, so i get 500m. But the answer states 714m. Where have i possibly went wrong?
 
  • #9
Kurdt
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No, the initial vertical component is not 50, is it? :smie:
Thats where you went wrong. I missed that :redface:
 
  • #10
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I'm sorry. But my physics teacher did not explain it clearly (even after i asked her). I can't really understand the book. Its not CLEARLY stated out. Someone mind to guide me? for the vertical component, i just calculate everything in the vertical direction? but in the question, the only initial velocity given is 50m s-1 no?
 
Last edited:
  • #11
Kurdt
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When the plane first drops the parcel, it has no vertical velocity since the plane is traveling at a constant height. Once the parcel is released its speed increases from 0 to its final speed because it is being accelerated by gravity. So initial vertical speed is 0m/s.
 
  • #12
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thanks for the reply. Even with that
1000m = (0)t + 5t^2
t=14.142s

then i used
u(t) = s

50(14.142) =707.10

still not the same as the book's answer.
 
  • #13
Kurdt
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Use 9.81 ms-2 for the acceleration due to gravity.
 
  • #14
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Thank you very much. What carelessness i've made.

what bout the second question?
Question 2:
When an aircraft is 4000m above the ground and flying upwards with a velocity of 50 m s-1 at an angle of 30 degree to the horizontal. A bomb is released. Neglecting air resistance, calculate the time taken by the bomb to reach the ground
(Assume g = 10 m s-2)

After solving the first question, i tried using the same method on my second question.
Firstly by using the vertical component
4000m = (0)t + 1/2(10)t^2
then i found t=28.28s

but the answer says the time is 31 seconds.
 
  • #15
Kurdt
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This time the initial vertical component of velocity is not zero is it since the plane is traveling up at an angle.
 
  • #16
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so first i find the vertical component with 50(tan 30) = initial vertical component of velocity ?
 
  • #17
Kurdt
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Yes almost. I wouldn't use tan though.
 
  • #18
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why? wouldn't tangent be easier since you have both adjacent and opposite value but don't have the hypotenuse ? I've tried substituting the value into it.

4000m = (28.87)t + 5t^2
t = -31 or 25 (31 is there, but how can it be a negative value?)
 
  • #19
Kurdt
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The velocity is the hypotenuse. You have to be careful with signs as well. You have defined downwards as your positive, so since the parcel will be initially traveling upwards you need a minus sign in front of its initial velocity.
 
Last edited:
  • #20
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Oh i totally forgot that velocity is going diagonally according to the angle, i've been thinking its traveling in the horizontal way. Thanks too for the reminder of signs, i've been careless.

4000m = -25t + 5t²
t = 31 or t = -25. Thanks!
 
  • #21
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sorry Kurdt i have another question here, related to the previous question too.

it asked for the velocity of the bomb when it strikes the ground. i used the formula v = u + at. Since i have u (-25) , a (g=10m s-2) and t = 31. Is the the right thing i should do?
 
  • #22
Kurdt
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Yes, but remember that the bomb will have a horizontal component of velocity aswell.
 
  • #23
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okay... what am i suppose to do with the horizontal component of velocity ?
I thought of finding the displacement with R = (u² sin 2Q)/g which is

R = (50² sin60)/10
=216.5m

then use
216.5 = 50t + 5t²
find the t and use
v= u + at

then merge the v = u + at of the vertical with the horizontal?
 
  • #24
Kurdt
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Horizontal is easy, its just the plane's component of velocity in the horizontal direction. Since, there is no air resistance that will be the answer.
 
  • #25
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sorry but could u explain that abit? so i just use the formula R = (50² sin60)/10 to find the distance ? That's all ?
 

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