Is the initial horizontial velocity the same as the starting velocity of a object for projectiles?

Question

An object is thrown 36.9 degrees above the horizontal, at a velocity of 15m/s, so isnt its horizontal initial velocity just 15? My professor said that they would be the same? But some people are saying to use trig to get the cos of 36.9 and then it would be 12...

Answer 1

You must have misunderstood your professor. If an object is thrown up at an angle greater than zero and less than 90 degrees both the horizontal and vertical components *must* be less than the combined velocity. Draw a right triangle where the hypoteneuse is the initial, angled velocity and you can see this immediately.

The reason the horizontal velocity is given by the cosine of the angle is that the *definition* of a cosine is the length of the side adjacent to the angle divided by the length of the hypoteneuse. Similarly, the vertical component is the side opposite the angle divided by the hypoteneuse, or the sine of the angle.

The sine of 36.9 deg is 0.600, which makes the (initial) vertical component equal to 9.00 m/s, and the cosine is 0.800, giving a (constant) horizontal component of 12.00 m/s.

Answer 2

You must have misunderstood your professor when he said that they would be the same.

What he meant, was that the x (horizontal) component of your initial velocity will always be the same, at all times. (during flight)

In projectile physics, there is almost always no horizontal acceleration, thus revealing the fact that the horizontal velocity of your initial velocity vector will always remain constant.

However, the vertical component of your initial velocity is subject to earths gravity, or -g. (-9.80 m/s^2; an negative acceleration)


To get the x (or y) component of your initial velocity, multiply the initial quantity by cos (or sin) of your projectile angle. (36.9)

Answer 3

The way the problem is stated, the initial horizontal velocity should be 12.0 m/sec.

The best way I can think of explaining it would be with a gun. For our purposes, regardless of the direction in which you fire a bullet, the bullet will eject with the same speed. (I'll make up a number, say 350 m/sec.) However will the bullet cover the same horizontal distance in the same time if I point the gun up at a 45-degree angle as it would if I pointed it perfectly horizontally? If the sun was shining straight down on horizontal ground and the bullets casted shadows (and if you could see in bullet time), you'd notice the shadow of the 45-degree bullet is traveling about cos(pi/4) times (~71%) the speed of the horizontal bullet (~250 m/sec in this case). This is even though they both left the barrel at 350 m/sec in their respective directions.

Answer 4

Total velocity in 2D (X&Y), has horizontal and vertical components.

"Some people" are correct in saying that the horizontal component is the cos of 36.9 multiplied by the total initial velocity.

Answer 5

The velocity along its path is 15m/s. The horizontal component of that velocity is 15*cos(36.9º).

Answer 6

nope, the initial horizontal vel is 15*cos(36.9)............u better dump ur professor who dun know anything

Answer 7

hi chum MuffinMommy, basically interior the previous artwork I even have given you the formulation to get the mandatory acceleration. wish you may persist with and do on your own. all the main suitable.