SOLUTION: Hi. I'm sorry to bother you but I just can't figure this question out and could really use your help. Here's the question: A cat jumps off a table which is 1.5 m high. If the i

Question 810869: Hi. I'm sorry to bother you but I just can't figure this question out and could really use your help. Here's the question:
A cat jumps off a table which is 1.5 m high. If the initial velocity of the cat is 10 m/s, at an angle of 37 degrees above the horizontal, how far from the edge of the table does the cat land?
My AP Physics teacher already gave me the answer for the problem, which is 11.5 meters but she wants me to show how I get to 11.5 meters by showing my work. I think I used all the right equations like v= vi + at to get the time and v= x/t for the distance but I still couldn't get to 11.5 meters. I think it has something to do with 1.5 meters. It must go somewhere in the equations but I can't figure out where.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let us find the time the cat is in the air
t seconds for total height
the total height reached y = y(0) +y(v)
time taken for y
1.5+10cos 37t+1/2(-9.8)t^2
but when it touches the ground y=0
4.9t^2-6.01t-1.5=0
Find the roots of the equation by quadratic formula
iiiii
a= 4.9 , b= -6.01 , c= -1.5

b^2-4ac= 36.12 + 29.4
b^2-4ac= 65.52

x1=( 6.01 + 8.09 )/ 9.8
x1= 1.44

x2=( 6.01 -8.09 ) / 9.8
x2= -0.21
Ignore negative value
t=1.44
d= rt
r= horizontal component 10 cos 37
d=10cos 37 * 1.44
d=11.50 meters

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