Physics problem. Help me out?

Question

The launching velocity of a projectile is 20 m/s at 53 degrees above the horizontal.

a.) What is the vertical component of its velocity at launch?

b.) Its horizontal component of velocity?

c.) Neglecting air resistance, which of these components remains constant throughout the flight path?

d.) Which of these components determines the projectile's time in the air?


I know this is a long one so if you can help me with any parts, preferably with an explanation rather than just an answer, I would really appreciate it. If you're looking for something to do with a head full of Physics knowledge, it would be pretty awesome if you emailed me.

Thanks!! :)

Answer 1

Draw a coordinate system, and draw a ray 53 degrees from the positive x axis. That is your vector. Using trig, you can draw x and y components for that vector, and solve for them using the length of the hypotenuse.



a) vert. component = x component = (Initial velocity) x (sin(initial angle of velocity))



b) horiz comp = y comp = (initial velocity) x (cos(initial angle of velocity))



c) Remember one thing: with projectile motion, the VERITCAL component of velocity changes because there is a force in the vertical direction, which cause ACCELERATION. that is the acceleration of gravity. There needs to be a force to cause a change in the component of velocity, meaning a force causes an acceleration. Now, with this information, realize that there is no force in the HORIZONTAL direction. Therefore, there is no acceleration in the horiz direction. And this the horiz comp of velocity does not change since there is no force and thus no accerlation. So the horiz component is constant!



d) Vertical. Because it's in the air until it hits the ground, and it hits the ground at a time T after is is thrown. Recall that horiz and vert motion are independent of each other. when asked as to how long it is in the air, that means how long until it hits the ground! so you really don't care about the horiz motion, because the vert motion and vert velocity is what decides WHEN the ball reaches the ground! also the formula v=v0 + at doesn't apply to horiz because the term "at" goes away due to the fact that a=0. So you can't determine time that way. You need to use the y component. Also think about this way: let's say you throw a ball up in the air straight, and you throw one at an angle from the horizontal. They will both hit the ground at the same time!



Wow the person below me claims to be a physics major, yet does not know the basics!

Answer 2

Ok!



So draw this out using a right triangle. The projectile starts on the path along the hypotenuse. Then the angle inside the triangle where the projectile launches is the 53 degree angle. The horizontal component is also along that side of the triangle and the horizontal component is of course the straight up side.



a) Vertical component (y): Using trig we see that the y-component is (20 m/s)*sin(53). About 15.97 m/s



b) Horizontal component (x): Again, use trig. (20 m/s)*cos(53). About 12.04 m/s.



You can see that that kind of makes sense because it is going faster upward than sideways. Since the angle is pointed above 45 degrees you can kind of imagine this like a roman candle pointed out of your hand.



c) The object will go up only for a while but then fall back down. Not constant!



d) It stays in the air until hits the ground! A hint for this is that the ground is where your vertical component basically doesn't exist.



Let me know if you need more help.

Answer 3

Assume that the projectile is fired from the origin of the coordinate axes. In this question you can resolve the velocity vector into two components one along the x axis and the other along the y axis(I hope u know resolution of vectors coz without that it is not possible to do the chapter of projectile motion). The one along the y axis is your vertical component and is = 20 * sin53 (please draw it and see) and the horizontal component is 20* cos53.



About the 3rd part notice that the vertical component of velocity is affected by the acceleration due to gravity (g=9.8m/s2 acting downwards) i.e. as the particle goes up it is decelerated and as it comes down it accelerates. the horizontal component is unaffected as no acceleration in horizontal direction is possible.



In the 4th part the horizontal component and the vertical component both can be used. Let R be the horizontal displacement of the projectile and let H be the max height attained by the projectile, T is the time of flight.



T=R/v(hor)

H=v(ver)(T/2)-0.5g(T/2)^2



But if u dont know H or R then only the vertical component can be used.



y= vsintheta*T-0.5gT^2, y=0 at the landing point of projectile. Now solve.