Projectile motion, vector components

[shawli]

Homework Statement

A child throws a snowball with a horizontal velocity of 18m/s directly toward a tree, from a distance of 9.0m and a height above the ground of 1.5m.

a) After what time interval does the snowball hit the tree?
b) At what height above the ground does the snowball hit the tree?

Homework Equations

displacement on x = horizontal velocity * time

displacement on y = initial vertical velocity * time + 0.5 * acceleration * time squared

or

displacement on y = final vertical velocity * time - 0.5 * acceleration * time squared

The Attempt at a Solution

For a), using the first formula, the calculation is:
9.0 m = 18m/s * t
t= 0.5s

Now I'm not sure what to do with b). If I use the equation:

displacement on y = initial vertical velocity * time + 0.5 * acceleration * time squared

is the initial vertical velocity 0? Since the boy throws the snowball horizontally? I think it is, but I don't get the right answer when I do this (the answer is 0.3 m).

What I'm doing is:

displacement on y = (0) * (0.5) + 0.5 * (-9.8) * (0.5)^2

 

[Karmalo]

"a" answer is correct

for "b" you can use:
$$Y=Y_{0}+V_{0}t+\frac{1}{2}at^{2}$$
where Y is the final altitude, $$Y_{0}$$ equals the initial altitude (1.5m), a the acceleration (a=g=-9,8(m/s^2)) and t is the time.

you resolve that and you'll get Y=0,275 (m)

 

[Doc Al]

What I'm doing is:

displacement on y = (0) * (0.5) + 0.5 * (-9.8) * (0.5)^2

That will give you the displacement from the original height of the snowball when it was thrown. How high was that above the ground?