Two students are passing a ball back and forth, allowing it to bounce once between them. If one students bounce passes the ball from a height of 1.4 m and it bounces 3 m away from the students, where should the second student stand to catch the ball at a height of 1.2 m? Assume the path of the ball is linear over this short distance.
So I have to answer this for homework using trig functions and stuff, but I'm pretty sure I can put it into a propotion
Since the angle of incidence is the same as the angle of reflection, we have two similar triangles. Thus, the distance d from the bounce to the second student is found by
d/1.2 = 3/1.4
Quick question, how would I solve this through trig
basically the same way. You have an angle θ, such that
tanθ = 1.4/3
Then, using the same angle,
tanθ = 1.2/d
you can either solve for θ, or eliminate it, producing the proportion used.
extra credit: use the more accurate parabolic trajectories to arrive at your answer.
So, the real question here is -- how close to a straight line are the bounces, really? I suspect not very close, depending on the speed with which the ball was first thrown.
Quick question, how would I solve this through trig
To solve this problem using proportions, we can set up a simple ratio between the heights and distances of the bounces. Let's assume that the distance between the two students is x.
We can compare the ratio of the heights of the bounces to the ratio of the distances of the bounces:
(height of second bounce) / (height of first bounce) = (distance of second bounce) / (distance of first bounce)
Substituting the given values, we have:
1.2 m / 1.4 m = x / 3 m
To solve for x (the distance between the students), we can cross-multiply:
1.2 m * 3 m = 1.4 m * x
3.6 = 1.4x
Divide both sides of the equation by 1.4:
x = 3.6 / 1.4 ≈ 2.57 m
Therefore, the second student should stand approximately 2.57 meters away from the first student to catch the ball at a height of 1.2 meters.