A projectile is launched at a given angle and hits the ground at a point 2m lower...

Question

A projectile is launched at a given angle and hits the ground at a point 2m lower than that of the origin. The maximum height of its flight is 3m and is attained 8m away from the origin. Which of the following most nearly gives the value of the total horizontal distance traveled (R).

Answer: R = 18.3

Answer 1

To solve this question, draw a free body diagram. It's going to help!

Draw a cliff with a 2-meter drop and draw the max height as 3 meters above the plane where the person has thrown the projectile. This will make the total height at the max point become 5 meters. This is important to grasp in order to get the question right because the question states that the max height is 3 meters only relative to the person who threw the object.

Now that we got that out of the way, we will need to break this problem up into 3 parts: 1) The y-component to figure out the time of flight before the max height, 2) the y-component to figure out the time of flight after the max height, and 3) the velocity of the x component.

Let's start off with 1)

1) Finding the time of flight before the max height:

(V )² = (V_oy )² + (2)(a)(delta y)

Velocity (V) at max height is 0 m/s at the y axis and we will plug in 3 meters as the max height because this is the max height of the rock being thrown off the cliff. We are already on top of the cliff when the throwing happens, so the max height when the rock goes up is just 3 meters for now.

0 m/s = (V_o )² + (2)(-10 m/s²)(3 meters)

V_oy = sqrt60 m/s = 7.746 m/s

Now that we have found the initial velocity of the y-component, we will use it to find the time of flight up until the max height.

V = V_oy + (a)(t)

0 m/s = 7.746 m/s + (-10)(t)

t = 0.7745 s

2) Finding the time of flight after the max height:

y = 1/2(a)(t²) + (V_oy )(t)

In this case, V_oy will be 0 m/s, because once the object reaches max height, the object will no longer have a vertical velocity. Thus, we can simplify the equation to this but only under the condition that we have reached max height!

y = 1/2(a)(t²) + (0 m/s )(t)

y = 1/2(a)(t²)

Now, for our height, we will plug in 5 meters because now we care about how long it will take the rock not only to fall the 3 meters it flew up back to neutral level, but also an additional 2 meters to below the cliff.

-5 = 1/2(-10)(t²)

t = 1.00 s

Now that we have found the time of flight before the max height and after the max height, we will add them together and have a total time of flight of 1.7745 s.

3)

Now, we deal with the x-component, which means that we need to find the velocity of the rock in the x-direction. Luckily, we know from the question, that it traveled 8 meters horizontally to reach its max height, so we will need to divide 8 meters by the time the rock needed to take to reach its max height when it was thrown, which is 0.7745 s.

V_ox = 8 m/0.7745 s = 10.32 m/s

Finally, to obtain the total distance traveled by the rock when it was thrown off the cliff, we will need to multiply V_ox by the total time of flight.

V_ox x total time of flight = total distance

10.32 m/s x 1.7745 s = 18.31 meters total distance.

I hope that this has helped you with any confusion with this question. Physics questions like these are best one by breaking down the problem in smaller more doable parts. If you need more help, feel free to visit my page and book a tutoring session. I would be happy to help you further!