002 (part 1 of 4) 10 points FOR PROBLEM 5, BE SURE TO MAKE YOUR ANSWER NEGATIVE. A ball is thrown horizontally from the top of a building 44.3 m high. The ball strikes the ground at a point 33 m from the base of the building. The acceleration of gravity is 9.8 m/s 2 . Find the time the ball is in motion. Answer in units of s.
003 (part 2 of 4) 10.0 points Find the initial velocity of the ball. Answer in units of m/s.
004 (part 3 of 4) 10.0 points Find the x component of its velocity just before it strikes the ground. Answer in units of m/s.
005 (part 4 of 4) 10.0 points Find the y component of its velocity just before it strikes the ground. Answer in units of m/s.
Hello notrandel! Haven't seen a physics problem for a long time. I might be a little rusty, so be sure to check my work before submitting!
Problem 002 (Where is 001? Or did you already finish that?)
First, let's draw a Free Body Diagram. Always do this at the beginning of a problem. (I'm not doing it here for the sake of time, but you can draw one on your paper.) The ball is thrown horizontally, so there is no initial vertical motion. So, we know that $v_{yi}=0$. We also have our vertical displacement, -44.3, and also the acceleration of gravity, -9.8 (not putting units for the sake of neatness). We can plug this into our 2nd Kinematic Equation, or $\Delta x=v_0 t+\dfrac{1}{2}at^2$, which gives, using the quadratic formula to solve for $t$, 3.01s. We can eliminate the negative solution since it doesn't make sence in the context of the problem.
Problem 003
Looking back at our previous solution, we find that the time it takes for the ball to hit the ground is 3.01 seconds. Since we know Range is the initial x velocity times time (I'm assuming air resistance can be neglected), we can plug our values in to the equation, getting $v_{xi}=\dfrac{R}{t}=\dfrac{33m}{3.01s}=10.96m/s$.
Problem 005.
I think you really should try this one yourself before checking your work with my solution. These should be simple now that you've read my previous solutions.
This problem is just plugging values into the 1st kinematic equation, or $v_f - v_i = at$, which gives us $v_f=-29.47$
Please tell me if I got these correct!
Hope this helped!