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Physics: Post your doubts here!

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Awesome12

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Mahnoorfatima said:
Physicsclassroom.com
A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper. Could you please solve this one as well. I don't GET this at all! The same source.
Click to expand...
Are the answers :
t = 1.15 s
Horizontal distance = 12.2 m
Peak height: 9.1 m
 
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Awesome12

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Mahnoorfatima said:
Last one's 1.6. First two are right. Can you please explain?
Click to expand...
Yeah sorry. A minor calculation error :D

Ok so basically a long jumper's jump's shape is that of a parabola, that is the same shape of the line of motion of an object in a projectile.

Form a right angle triangle whose angle with the horizontal is 28 degrees and whose hypotenuse is 12 (We are indirectly splitting the motion of the object in to horizontal and vertical components)



I)

Now the first question requires us to find the time the body is in the air. So we have to decide which component to use: Horizontal or Vertical?

Acceleration is a common term in each of the 4 or 5 formulas of motion. Acceleration is basically the force of gravity (9.81 m/s^2).

Since the body is moving upwards, the acceleration will be negative. (ie. -9.81 m/s^2).

Acceleration is a vertical component since gravity acts downwards. Therefore whatever formula we use, we have to use the vertical components only.

Now we know that the displacement is 0, since the body moves upwards and then returns to a level equal to the mean position.

Now find the VERTICAL COMPONENT of velocity. Why? Because like I said before, all terms should be the vertical components.
VERTICAL COMPONENT OF VELOCITY = 5.63 m/s


So far what we know:
Acceleration = -9.81
S/Displacement = 0
VERTICAL COMPONENT OF VELOCITY = 5.63 m/s

Time = ?


Now find a formula whose terms contains the things we know

One formula is

s = ut + 1/2 at^2

Substitute all the values to get t in the form of a QUADRATIC EQUATION:

Solve the quadrative equation. You wil get 1.15 and 0

Since the time cannot be 0, the ANSWER IS 1.15




II)

Now in part II, we need to find the HORIZONTAL DISTANCE. We already know the time.

What we can also derive is the HORIZONTAL COMPONENT OF SPEED, as like I said before, all the components should be the same either VERTICAL or Horizontal.

HORIZONTAL COMPONENT OF SPEED = 12 (Cos 28) = 10.6 m/s

Distance = 10.6 * 1.15

Distance = 12.2 m




III)

Now we need to find the peak position (that is the maximum VERTICAL DISTANCE)

Again like I said, to find the VERTICAL DISTANCE, the rest of the terms should also be VERTICAL COMPONENTS

We already know the VERTICAL COMPONENT OF INITIAL VELOCITY = 5.63 m/s

We also know the FINAL VELOCITY = 0
Why is it 0? Well, at the top, there is no velocity whatsoever, so we consider it to be 0.


What do we know so far

VERTICAL COMPONENT OF INITIAL VELOCITY = 5.63 m/s

FINAL VELOCITY = 0

ACCELERATION = -9.81 m/s

We need to find s/Displacement = ?


Formula to be used:

v^2 - u^2 = 2as

Substitute all the values and find s

s = 1.6 m



I hope this helps. :)
 
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bludlynsolja

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thx a lot physicist for the link, it really will be useful!!!
not everyone out there would be wiling to share such resources, thx a lot bruh!
have similar stuff for chemistry? i know this is the physics section bt just inquiring whether u have stuff like that for chem?
thx once again :)
 
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bludlynsolja

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ok for part b, u have to find the speed of the bullet before hitting the block, so u have to apply rule of conservation of momentum which, if i paraphrase, states that the total momentum before collision must equal total momentum after collision in a closed system, hence:
momentum before collision: mass of bullet* velocity of bullet + mass of block* velocity of block.
so we change masses to the right SI unit in this case kg and remember, the block won't have momentum initially as it is at rest so 0 velocity before the collision.
so momentum before collision will be momentum of only the bullet as its fired so mv= 0.002kg* v
momentum after collision would be the total mass of both bodies times their velocity, bcd, remember they stuck together as u are told in the question so u treat it as one mass in that case, so total mass will be (0.6+0.002)= 0.602kg and velocity is 1.3 as told in the previous part
so if we equate the mv before collision and after, we get:
0.002v=0.602*1.3, and hence, we get v as 391mtrs per sec.

c.) here u have to find kinetic energy of the bullet before impact, and k. energy is given by 1/2mv^2 and we already found the velocity as 391 so just a matter of substituting so 1/2*0.002*391*391=153 joules

the second part is asking to state the type of collision, u should know that there is two types of collision, the inelastic collision and elastic collision. inelastic is where kinetic energy is not conserved that is, it is different before and after the collision while elastic is where the kinetic energy remains exactly the same before and after collision so in this case, if u find the kinetic energy after collision it will be 1/2*0.602*1.3^2= 0.5 joules so clearly u can see that the kinetic energy before was 153 and after was 0.5 so its not the same so it is inelastic.
 
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bludlynsolja

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http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Physics (9702)/9702_w03_qp_2.pdf
ok so for question one part b 2, the first one is after 0.7 sec right, so here u just have to count the number of seven successive dots starting from the first one, coz u are told each dot is a flash of the object after 0.1 sec so for 0.7 sec it will be 0.1*7 which is 0.7 sec, so u count upto the 7th dot and read the value given and u will find it to be 132 cm.
now the one for after 1.1 seconds is abit challenging, check eh:
if u notice the trend, the thing starts moving with constant speed, how do i know that? well if u see the spacing of the ball, it becomes the same spacing after
around the fourth photograph, so this same spacing means that the distance covered in every 0.1 sec is the same and since distance is the same and the time is also the same i.e every 0.1 sec, so u can definitely tell tat the speed is constant since speed is distance/time. am trying to go step by step so u get it....
so now, what is this distance? well if u see, the distance is 25 meters from each flash starting the fourth photograph, for instance, take the fourth photograph, what distance can u read? u get 58cm, then take the fifth photo, the distance is 82 s if u minus 58 from 82 u get the distance as 25cm, then take the sixth photo u see, the distance is 108 and the fifth one is 82 so if u minus again, u still get 25 cm, so this shows that now the distance after each photograph will have a difference of 25cm from the preceding phograph so if we understand that concept, then we can find the distance for 1.1 sec as follows:
we know that till 0.7 sec the distance is 132 cm as calculated earlier and now we know that for 0.8 sec the distance will be 132cm + 25= 157, then for 0.9, it will be 157+25= 182cm then for 10.0sec it will be agin 182 plus 25 u get, 207cm and finally for 1.1, it will be 207cm plus 25 and u get 232xm and thats ur answer....

c ok for part c we have to find the number of photographs when it is a lead sphere, here we need to first use the equations of linear motion and i am gong to use s=ut+1/2at^2 and we know it is falling under gravity so the value for acceleration will be 9.8ms^-2, i hope u r familiar with this idea of gravity, if not just know that when a body is falling vertically =, then it falls under the influence of gravity and the value of acceleration is hence 9.8 or some books give 10 so using the equation now.
we know the distance is 160cm as told in the question, so in meters thats 1.6m, the value of u, the initial velocity is 0 as it started from rest.
so now we have: s=ut+0.5at^2 , hence, 1.6= 0t + 0.5*9.8*t^2 so if we now find t, we will get it to be 0.57seconds,
but we are told to find the number of photographs, but we know that each photo is after 0.1 seconds so
0.1sec= 1 photograph
0.57sec= x photogrpahs, if we cross multiply, we get 5.7 photographs.
bt we can't have a fraction of a photograph if we think of it practically so we take the next number which is 6 so photographs.
 
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