Elementary mechanics

Question

an object is shot from the ground at 125m/s at an angle 30o above the horizontal. how far away does the object lands

Answer 1

Let's first determine how long the object is in the air for. To do this we'll need to know it's initial y velocity.

The y component of our velocity is vy=vsin(b) where vy is the y velocity v is the speed it is shot and b is the angle with respect to the horizontal. This means our vertical velocity is 62.5 m/s.

We'll need to following equation to determine how long the object will be in the air.

xf-xi=vi*t+.5 at^2

where xf and xi are final and initial positions, vi is the initial velocity, a is the acceleration and t is time.

Because the object starts and ground level and will end at ground level the change in height is zero. We have a positive velocity going up at 62.5 m/s and a negative acceleration of 9.8 m/s. Plugging things in

0=62.5*t+.5*9.8*t^2

solving for t we get

t=0 and t=12.76 seconds

The t=0 solution is when we initially shoot the object with the 12.76 seconds being the time that the object hits the ground after being launched.

Now we'll need to determine the horizontal velocity. Using the equation vx=vcos(b) we find that the horizontal velocity is 108.25 m/s. With no acceleration being applied in the horizontal direction we can use the above equation again to determine how far the object travels.

xf-xi=vi*t

If we set our intial position xi at 0 we get xf=108.25*12.76 which is 1381.31 m.

Answer 2

The question is asking for the range of the projectile. The initial horizontal velocity of the object is 125 m/s cos(30º) = 125√3/2 m/s. If t denotes the object's flight time, then its range is (125√3/2)t. If t* is the time at which the object reaches max height, then t = 2t*. Hence, the range is (125√3/2)(2t*) = (125√3)t*. To find t*, note that at max height the object's vertical velocity is zero. Therefore, the initial vertical velocity equals gt*, or 125 sin(30º) = gt*. That is, 125/2 = gt*. Dividing by g we obtain t* = 125/(2g) and consequently the range of the object is (125√3/2)(125/(2g)) = 15625√3/(4g) meters, which is approximately 690 m.