2-Dimensional Motion

Question

A hiker walking along a trail on a 300 meter cliff encounters an interesting looking rock. He promptly kicks it off the cliff with a velocity of 20 m/s with a 15-degree angle. How far away does the rock land?

I am kind of confused with the whole concept of the problem, are there three different times we have to solve?

Answer 1

Galena

I assume the 15 degree angle is the angle of elevation ??.

In which case, when he kicks it horizontal velocity = 20 cos 15° (this will be constant)

vertical velocity = 20 sin 15° (this will decrease because of gravity).

Assume positive direction is upwards and to the right.

Let s the vertical distance from the top of the cliff. So when t = 0, s = 0..

What is t when the rock hits the sea.

Well s = ut + 0.5 a t²

s = t. 20 sin 15° - 0.5 x 9.8 x t²

It hits the sea when s = - 300

So solve this quadratic for t. (plot in on a graphing calculator and find the zero).

Knowing t calculate the horizontal distance travelled 20 cos 15° . t

Then knowing the horizontal distance travelled when it hits the sea and the vertical distance fallen (300) you can find the distance travelled (use Pythagoras).

I get t is about 9.36 s, so horizontal distance is about 180.82 , vertical 300, so straight line distance to where it hits the sea is 350.28

Answer 2

The way to deal with these problems is to break them down into what is happening in the x-direction and what is happening in the y-direction and handle those as separate problems. So, break down the initial velocity (20 m/s at 15° into v_x and v_y:

Using trig, we can say that cos(15°) = adj/hyp = v_x/20 so v_x = 20cos(15°) = 19.3185 m/s

Using trig, we can also say that sin(15°) = opp/hyp = v_y/20 so v_y = 20sin(15°) = 5.1764 m/s

So, we are being asked how far away the rock lands which is the distance in the x-direction. The only force that the rock has is the force of gravity which is acting in the y-direction so there is no force in the x-direction that will change the velocity v_x. That means, considering ONLY the x-direction, the rock will be traveling at 19.3185 m/s until it hits the ground. We can say the distance traveled is the velocity times the time. But we don't know the time. So, let's try to determine the time using the y-direction.

In the y-direction, the initial velocity is 5.1764 m/s. The acceleration due to gravity is 9.81 m/s² so we can use the equation x = v_it + 1/2at² to solve for time (t).

300 = 5.1764t + 1/2(9.81)t² (I'm being a bit lazy here in defining the downward direction as positive making the 300 a positive, the 5.1764 a positive, and the 9.81 a positive. It might be more traditional to define up as positive which means all those numbers would be negative but can you see that it won't change the math?)

We solve quadratics by using the quadratic formula. Set the equation equal to zero and simplify a little to get:

4.905t² + 5.1764t - 300 = 0

So t = [-5.1764 ± √(5.1764² -4(4.905)(-300))]/(2•4.905)

t = [-5.1764 ± √5912.8]/9.81

t = (-5.1764 + 76.895)/9.81 and t = (-5.1764 - 76.895)/9.81

So t = 7.3107 s and t = -8.3661 s and we can throw out the negative answer.

So, now we have the time. Going back to the x-direction, distance = velocity•time = 19.3185•7.3107 = 141 meters. There is only 1 significant figure in the given numbers, so to be really correct, we would need to round to 100 m but I'm not sure how fussy your teacher is about that.

So the rock travels 300 m down and 141 meters in the x-direction. To find the distance it actually traveled along the arc is a lot more complicated but I doubt if that is what they are asking for.