# I need urgent physics help.My teacher didn't explain this well. Please help with as many as u wish but if i can just get help with one at least i can figure out how to do the rest. 1. An Olympic...

I need urgent physics help.

My teacher didn't explain this well. Please help with as many as u wish but if i can just get help with one at least i can figure out how to do the rest.

1. An Olympic long jumper leaves the ground at an angle of 23 degrees and travels through the air for a horizontal distance of 8.7m before landing. What is the takeoff speed of the jumper?

2. A driver springs upward from a board that is three meters above the water, A the instant she contacts the water ger speed is 8.90 m/s and her body makes an angle of 75.0degrees with respect to the horizontsl surface of the water. Dtermine her initial velocity, noth madnitude and direction.

3. A soccer player kicks the ball toward a goal that is 16.8m in front of him. The ball leaves his foot at a speed of 16.0m/s and at an angle of 28 degrees above the ground. Find the speed of the ball when the goalie catches it in fornt of the net.

4. The Javelin is launched at a speed of 29m/s at an angle of 36 degrees above the horizontal. A it travels upward, its velocity points aboce teh horizontal at an angle that decreases as time pases. how much time is required for the angle to be reduced from 36 degrees at launch to 18 degrees?

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I believe you need to understand the basic concept common to the four problems and a general method to solve them. This is what I intend to clarify in my answer here.

When any object is thrown in the air, the movement of that object can be resolved in two separate movements:

- Movement along a direction parallel to the earth, or any horizontal surface like a playground or surface of water in a swimming pool.
- Movement in a vertical direction.

Assuming that there is no air resistance to the movement to the object, the horizontal movement under the influence of initial horizontal velocity is uniform at that velocity.

However the vertical component of the velocity of vertical movement is affected by acceleration due to gravity, which on earth is equal to about 9.81 meters/second^2. Depending on initial direction in which the object is thrown the initial velocity vertical velocity can be in upward direction or downward direction. For objects thrown in a horizontal direction or just dropped from a height,this initial velocity can also be zero. A object with initial velocity in upward direction will start rising up, but with reducing velocity because of acceleration due to gravity. In this way the object will continue to rise till its velocity becomes zero. This is the highest point reached by the object. Immediately on reaching its highest point the object start falling down with increasing velocity. With the velocity at highest point being 0 (zero). The object continues to fall till it reaches ground or some other obstruction in its path.

Frequently the Object may be bounced off the surface it hits this way, and may continue to move further. However normally in problems of this type, designed to provide provide practice in use of principles of projectiles, the movement after the object hits the ground is not considered.

The exact path path of motion of the object is the combined effect of the vertical and horizontal movement. In science this path is sometimes called a projectile, which has the shape of a parabola.

In solving problems of this type the horizontal and vertical components of initial velocity can be determine by the following formulas:

Horizontal velocity = Actual velocity*Cos A

Vertical velocity = Actual velocity*Sign A

Where A is the angle that the initial velocity makes with the horizontal.

If horizontal velocity (H) and vertical velocity (V) are known the resultant velocity (R) can be calculated as:

R = (H^2 +V^2)^1/2

The tan of angle A of the resultant velocity can be calculated as:

Tan A = H/R

Further quantities as required by the problem can be calculated using Newton's law of motion and the following equations giving relationship between different variables involved.

v = u + a*t

Average velocity = (u + v)/t

s = u*t + 1/2[a*(t^2)]

v^2 = u ^2 = 2a*s

Where:

u = initial velocity

v = final velocity (after time t)

a = acceleration

f = m*a

In case of acceleration due to gravity the symbol g may be used instead of a.

t = time of movement

s = distance covered (in time t)

f = force applied to accelerate the object

m = bass of the object

1) Let u be the take of speed of the Olympic jumper. So the relevant horizontal and vertical components of velocities are: u cosx and usinx , x being the direction of the jumpwith the horizontal.

(usinx)t-(1/g)t^2=0 gives time he grounds: t=( 2usinx)/g

Therefore the horizontal distance he covered = (ucosx)t=(ucosx)(2usinx)/g equal to 8.7m solve for u, given x=23 deg.

u^2=(2.7)*9.81/ (2sin23cos23)= 18.4106m/s

2)let u be the take of speed of the diver. The initial horizontal and vertical components if velocities are : ucosx and usinx.The The divers speed at the time of contacting water is v = -sqrt[ (usinx)^2+2 (-g) (-3)]= -sqrt[(usinx)^2+6g] as the velocity is towards earth.

Therefore, (usinx)^2+6g = 8.9 ^2 ...................(1)

tan x= tan75 = vertical velocity compent/ horizontal velocity compent or

sqrt((usinx)^2+6g) / ucosx = tan75

(usinx)^2 +6g= (ucosx)^2*(tan75)^2.................(2)

From (1) and (2) you can determine u and also x.

3)The horizontal and vertical components of the ball are ucosx and usinx. u= 17, x=28 degree above horizontal.

Assume the catching and kicking is at the same level of horizontal plane where vertical displacement is zero.

(usinx)t-(1/2)g t^2 = s, s=0 when the ball is caught. So,

t= (2usinx)/g

Final vertical component velocity = initial vertical component of velocity -g*time

=usinx - g(2usinx)/g= -usin x

Horizontal component final velocity = ucosx.

Therefore, the final magnitude of velocity =sqrt{ (u cosx)^2+(-usinx)^2}= u= 16 m/s

Direction: Tan inverse (vertical velocity/ horizontal velocity )=

=tan inverse (-usinx/ucosx) = - x degree = -28 degree to horizontal . This idicates the magnitude of the velocity is below horiontal 28 deg.

4) vertical component of velocity/ horizontal component of velocity = tan 18 degree

(usinx-gt)/ucosx = tan18. Given u=29m/s and x= 36 degree solve for t.

I'm going to give you a general idea of how to approach these problems. When you are working in any science, but especially in physics, write down every piece of information given in the problem--there are usually only a few types of equations you are learning about at any one time. You then look at the values you have, compare them to the equations you are working on--and amazingly enough, you will usually have all the pieces of information for one type of problem, with a missing piece that will give you your answer. You may have to move the values around so that you get the one you are looking for by itself, but you can do it.

Now look at number 1. You have distance traveled = 8.7 m., angle of takeoff = 23 degrees, and you are looking for takeoff speed (initial velocity.) You may have to use your trig (tangents, sines, and cosines), you may have to use acceleration due to gravity, you may have to draw a picture (in fact, you probably should),to do a step before hand, but is there a formula you have that will work? Look back at the examples given in class, or in your book, and you will find something similar... Write down what you know. Find a formula that uses them. Draw a picture.

You can do this!