I have this question...
A girl throws a ball vertically upwards with a speed of 8 m/s from a window which is 6 m above the horizontal ground.
a) find the geatist hight reached by the ball.
I managd to do this question with an answer of 9.27m.
This is the question with wich i am struggaling...
1.5 s later she drops a second ball from rest out of the same window.
b) Find the distance below the window at which the balls meet.
First of all, use s = ut + [sup]1[/sup]/[sub]2[/sub]at[sup]2[/sup] to write down the two relevant equations of motion. I shall take "upwards" motion to have a positive speed:
Whereis the height of the ball above the window at time seconds after it is released, and is the height of the ball above the window at time seconds after it is released. It seems that the big step to solving this problem is seeing that and differ by 1.5 seconds, and we want that when . So:
Having established this, we substitute into one of our original equations of motion, giving:
Now, we want to find when- in other words, when the following occurs:
From this point, the rest of the working is just algebra to find the time at which this event occurs;
And you can use this value ofto get the distance required.
Is this all clear?
Bad speling makes me [sic]