Physics Homework Help
Can anyone on Basil explain how to solve this? I just don't get it
A ball is thrown from the top of a 12m cliff with an initial velocity of 36m/s at an angle of 40degrees above the horizontal.
a) How far from the base of the cliff does the ball land? (answer is 140m)
b) What is the velocity of the ball at the moment of impact? (answer is 39m/s, 45 degrees below horizontal)
Thanks.
July 6, 2012
5 Comments • Newest first
Oh ok, I get it now. Thanks guys!
@Fiercerain Hey Fierce! <3
draw a triangle with the ball, and you'll see that 36sin40º is the vertical speed of the ball, and that 36cos40º is the horizontal speed. Then, using the formula d=v1� t + a(� t^2)/2, you can find the time it takes (vertically) for the ball to hit the ground using 36cos40º as v1 and a as either 9.8 or 10, and displacement as the height of the cliff, 12m. Using that time it takes for the ball to hit the ground, you know that there is negligible air resistance horizontally, so take the formula distance=speed * time, to find the distance it lands from the base of the cliff. Finally, because you have the horizontal velocity and it doesn't change (36cos40º), draw a triangle with that velocity and the vertical velocity of the ball as it hits the ground (v2) using a formula with known variables and v2. Then finish it off by forming a triangle with these two vectors (36cos40º and v2) and find the hypotenuse. That's the velocity of the ball as it hits the ground (use trig to find the angle). Hope this helped
I wish I still retained what little I understood of physics. Hi there.
[url=http://s908.photobucket.com/albums/ac285/chao195/?action=view&current=physics_help.png]I was able to figure some of it out[/url]
idk how to expolain how to do it in a wall of text...
wait i got an idea ill make a picture so look back in a few minutes