Need help with this question!

[t5am94]

Alright this is from the 2007 Unit 3 Physics Exam. I worked the first two parts of the question, however i got stuck on the last one (which is highlighted in bold). Hopefully one of you guys can help me out.

Cheers.

A rock of mass 0.50kg is set on the ground, pointing VERTICALLY up, as shown in Figure 3. When ignited, the gunpowder burns for a period of 1.5s and provides a constant force of 22N. The mass of the gunpowder is very small component to the mass of the rocket, and can be ignored. The effects of air resistance can also be ignored.

What is the magnitude of resultant force? 17N

After 1.5s, what is the height of the rocket above the ground? 35m

A second rocket, that again provides a constant force of 22N for 1.5 s, is now launched HORIZONTALLY from top of a 50m tall building. Assume that in its subsequent motion the rocket always points horizontally.

Q: After 1.5s what is the speed of the rocket, and at what angle is the rocket moving relative to the ground?

[soopertaco]

I have no idea if this is correct but:
Fnet = 22 = (0.5)(a)
a = 44 m/s^2

so you have:
initial vertical velocity = 0 m/s
acceleration: 44 m/s^2
time = 1.5 seconds
v = u + at
v = 0 + (44)(1.5) = 66 m/s
haha im stumped, probably would do better if i had a diagram.

p.s you're from parade college, bundoora??

[thatricksta]

To get the angle, use a diagram.

Soopertaco has given you the velocity parallel to the ground.

but the question is quite ambiguous... it says "assume that in this subsequent motin the rocket ALWAYS points horizontally" =/

angle is 0 degrees?

[ttn]

You've got to calculate the vertical velocity at 1.5s as well.

v = u + at
v = 0 -10(1.5)
v = -15m/s

Draw triangle.
so speed becomes 67.7 m/s
angle = arctan(15/66) = 12.8 degrees to the horizontal.