Projectile Motion?

Question

A shell is fired forom the ground with an initial speed of 1700 m/s at an initial angle of 58 degrees to the horizontal
The acceleration of gravity is 9.81 m/s2
Neglecting air rexestance how do I find the shell's horizontal range and the amount of time the shell was in motion?

Answer 1

yfinal = yinitial =0 = Vy*t + 1/2 * (-g) * t^2

Vy = v * sin58 = 1700 * sin 58 = 1441.68

0 = 1441.68*t - 4.9*t^2

t(1441.68-4.9*t)=0

1441.68-4.9t=0

t = 1441.68/4.9

t=294.221 seconds = 4.9 minutes

range = Vx*t
range = 1700*cos58*294.221
= 265053m
~265 km

Answer 2

First, you need to solve for the object's initial horizontal and vertical speeds.

If it was fired at 1700 m/s at an angle of 58 degrees to the horizontal, then you know that its horizontal speed is:

(1700 m/s)*cos(58 degrees) = 900.863 m/s

And you know the vertical speed is:

(1700 m/s)*sin(58 degrees) = 1441.68 m/s

In other words, the speeds form a triangle with the horizontal speed on one leg and the vertical speed on the other. The total speed is the hypoteneus. Notice how 900.863^2 + 1441.68^2 = 1700^2.

Now that you have that, you need to figure out how long the projectile will be in the air. In other words, you need to figure out how long it will take for it to shoot up, slow down, and fall back to earth and hit the ground.

The vertical position of an upward moving object is:

0.5*a*t^2 + v0*t + x0

Where a is the acceleration (-9.81 m/s/s), v0 is the initial velocity (1441.68 m/s), and x0 is the initial position (0 m). You are solving for the time (t) where that expression is 0. In other words, you need to solve:

-0.5*(9.81 m/s/s)*t^2 + (1441.68 m/s)*t = (0 m)

for t. It's clear that one solution to this problem is t=0 seconds. This is the trivial solution. It is at 0 m right when it gets shot up. Disregard this solution and divide by t. You then get:

-0.5*(9.81 m/s/s)*t + (1441.68 m/s) = (0 m/s)

Which gives you:

t = (1441.68 m/s)/(0.5*9.81 m/s/s)

which is:

t = 292.921 seconds

If you want, plug that into the position formula above and you'll see that you get 0.

So you know the projectile will be in the air for 292.921 seconds. Now all you have to do is find out how far it traveled horizontally. You can do that by multiplying the horizontal speed by the time it was in the air:

(900.863 m/s)*(292.921 s) = 264,782 meters

So the projectile traveled 264,782 meters and was in the air 292.921 seconds.

Answer 3

The shell will describe a parabola in a uniform gravitational field.

Discounting air resistance and coriolis force and assuming a completely flat landscape - use the suvat equations to find the time the shell was in the air and then find the distance it travelled over that time.

Okay - first you'll need to do some trig

..../
../
/58 deg
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the vertical velocity is given as 1700.sin(58) m/sec

the horizontal velocity is given as 1700.cos(58) m/sec

Use v = u + at to solve for t where v = 1700sin(58) and u = 0 and a is due to gravity. Double it cos the shell has to come back down.

Now you know t

you can find s, but bear in mind you are wanting to find HORIZONTAL range so use the horizontal velocity when you do this




Funny - I get more like 132km and 294 sec

Guess you'll just have to pick from the bunch of random answers you're getting :)

Duh - u(horiz) is also 1700cos58

Hence s = 265km

Answer 4

Use trigonometry to determine the horizontal and vertical components of the initial velocity. Then use v = v0 + at, with v = 0, to find the time to reach the top of the arc. Double that time to find the time until the shell hits the ground. Finally, use the total time of flight times the horizontal component of the velocity to find the horizontal distance (range) travelled.

Answer 5

Well first you find the component of the velocity that is upwards.
1700 * sin(58degrees) = v0.
1700 * cos(58degrees) = Vx.

y(t) = 0 = y0 + v0*t - 0.5 * g * t * t
solve for t given that y0 = 0

t= 293.92s
range = Vx * t = 264 782.344 m