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.

 

[baba89]

Your approach is on the right track, but there are a few things you can improve upon. First, make sure you are using the correct equations for motion in two dimensions, which are:

x = x0 + v0xt + (1/2)axt^2
y = y0 + v0yt + (1/2)ayt^2
vxf = v0x + axt
vyf = v0y + ayt

where x and y represent position, x0 and y0 represent initial position, v0x and v0y represent initial velocities in the x and y directions, ax and ay represent accelerations in the x and y directions, and t represents time.

For part (a), you will need to use the first equation, setting y0 = 0 since the ball is tossed from the edge of the building. You know the initial velocity in the x direction (v0x) and the time (3s), so you can solve for x.

For part (b), you can use the second equation, setting x0 = 0 and solving for y0. This will give you the height from which the ball was thrown.

For part (c), you can use the second equation again, but this time set y = -10m (since the point is 10m below the level of launching) and solve for t. This will give you the time it takes for the ball to reach that point.

Remember to pay attention to units and use the appropriate values for acceleration due to gravity (9.8m/s^2) and the angle (20 degrees) in your calculations.