You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity v at an angle è with respect to the horizontal. Let the building be 50.0 m tall, the initial horizontal velocity be 9.00 m/s, and the initial vertical velocity be 12.0 m/s. Choose your coordinates such that the positive y-axis is upward, and the x-axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ball’s maximum height above the ground, and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.
Could someone please help me get through this problem step by step? Even if its just hints that would be great. Thank u so much.
6 answers
è is theta. Sorry
I am not certain of your difficulty, if you sketch your coordinate system.
You have the initial velocities, and initial position. Initial position in a) is hyinitial=0, and if it falls, you should get a final h negative (it goes down).
hfinal=hyinitial + vvertialinitioal*time -9.8t^2/2
You have the initial velocities, and initial position. Initial position in a) is hyinitial=0, and if it falls, you should get a final h negative (it goes down).
hfinal=hyinitial + vvertialinitioal*time -9.8t^2/2
how can i find t then?
Hfinal is 50 m below the origin. solve for t.
so is it like this:
-50m = 0 + 12(t)-(9.8t^2)/2
-50m = 0 + 12(t)-(9.8t^2)/2
Could u write it step by step for me, please? It would be very helpful and would save me alot of confusion since my teacher has not taught us any of this yet. But we're suppose to do this problem. Thanks
2 answers