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
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
5 answers
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.
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.
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