Question

Help please! I will give a medal to the best answer!
A cat pushes a ball from a 7.00 m high window, giving it a horizontal velocity of 0.20 m/s. How far from the base of the building does the ball land?
(Assume no air resistance and that ay=g=9.81 m/s^2.)
A. 0.24 m
B. 1.4 m

Answer 1

Am I supposed to use the formula: v(y,f)=a(y)Δt ?

Answer 2

The variables in parentheses are subtitles, by the way. In my original post, ay is a subtitle too.

Answer 3

The question gives you information about VERTICAL stuff (e.g. 7 m is a vertical height), but wants you want to calculate a HORIZONTAL distance, so somehow you need to get from vertical to horizontal.

The way you do this is by looking at TIME, because the length of time the ball is traveling vertically is the same as the length of time the ball is traveling horizontally. If you can find this time from the vertical information, you can then use it to calculate the horizontal distance traveled.

In the vertical direction, we know d(y)=7, a=9.81, and v(y,i)=0 m/s. We want to find t. The equation that uses all these variables is the following:

\[d_y=v_it+frac{1}{2}at^2\]
After solving for t, you can then think about the horizontal distance. The ball has a constant horizontal speed, so you can just use v(x)=d(x)/t to calculate the range, d(x)!

Answer 4

Wow the equation got a little mixed up. Let me try again:

\[d_y=v_{(y,i)}t+\frac{1}{2}at^2\]

Answer 5

Okay, so once I plug those given numbers into that equation, it becomes:
\[7=0(t)+\frac{ 1 }{ 2}(9.81t)^{2}\]

Answer 6

@matt101
Is that right? How do I simplify it from this point?

Answer 7

0.24 m