Motion in 2 Dimensions homework problem =]

[riafran]

Homework Statement

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00m/2 at an angle if 20 degrees below the horizontal. It strikes the ground 3s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10m below the level of launching?

Homework Equations

not sure about these:

h=1/2gt^2
Vf - Vi = gt
Yf = Yi + Vi + (at^2)/2

The Attempt at a Solution

Well I started drawing a diagram with some point h on the y-axis and a line with a negative slope from some point h (on y-axis). This line hits x-axis at 3s, Vf = 0m/s. I was trying to find the height and used the above equation and got 44.1 and tried to find the horizontal distance.. i got 88.2 but i don't think that's right.. please help! thanks =]

 

[LowlyPion]

Homework Statement

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 8.00m/2 at an angle if 20 degrees below the horizontal. It strikes the ground 3s later.
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10m below the level of launching?

Homework Equations

not sure about these:

h=1/2gt^2
Vf - Vi = gt
Yf = Yi + Vi + (at^2)/2

The Attempt at a Solution

Well I started drawing a diagram with some point h on the y-axis and a line with a negative slope from some point h (on y-axis). This line hits x-axis at 3s, Vf = 0m/s. I was trying to find the height and used the above equation and got 44.1 and tried to find the horizontal distance.. i got 88.2 but i don't think that's right.. please help! thanks =]

The 44.1 only represents the effect of the 1/2gt². This equation is not complete. Your Y height also had the contribution of the initial downward y-velocity component.

That should give you the correct height.

Your distance is given merely by the x-component of the initial velocity times time. Starting at 8, and a component of 8 at that, and going 3 seconds puts anything over 24 as ... well suspicious.

To figure part C you use the first equation, but rather than solve for height, which they now give you, you must solve the resulting quadratic for time.