Thank you - this explanation really helped. I'll be changing this question to account for the actual horizontal displacement for the time that is given. In the meantime, I'll have to take both answers as long as the work for what was given is correct.
My students thank you :) This is what I get...
But it can't travel 70m horizontally because at the same vertical height on the other side is a cliff with a person on it that the arrow is hitting.
Are you telling me the angle is actually 2.34° and so my work finding my initial horizontal velocity is incorrect??I'm not sure this forum is...
This gives the incorrect initial velocity in the y for my angle that I determined using x-components.
Why is this the incorrect initial velocity in the y? That is my question.
At 0.75s it hits the other cliff (there is a person there that it hits). So yes, I know it's still high in the air, it's at the top of the other cliff.
I'm not sure I understand what you mean by "assuming an air time of 0.75s fixes the initial y-velocity, which in turn fixes the launch angle."
I considered Viy to be at the peak of the trajectory, considering the second half of the path. Knowing that my Vfy at the end of the path would be equal to the initial velocity at the beginning of the path.
Homework Statement
There are two cliffs separated by 40 meters (horizontal displacement). An arrow is shot from a bow at an angle and a velocity of 90 m/s. The arrow takes 0.75 seconds to arrive at the other cliff. What is the angle at which the arrow was released? Neglect air resistance...