Question

Exactly 3.0s after a projectile is fired into the air from the ground, it is observed to have a velocity v = (8.3 i^ + 4.9 j^)m/s, where the x axis is horizontal and the y axis is positive upward.

Determine the horizontal range of the projectile.

Determine its maximum height above the ground.

Determine the speed of motion just before the projectile strikes the ground.

Determine the angle of motion just before the projectile strikes the ground.

Answer 1

Can you write the equations of motion ??

Answer 2

You mean the components for Vx and Vy?

Answer 3

I mean these kind of equations -
http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

Answer 4

My problem is that I am confused as to which ones I should use. I know that I will probably need to use one equation to solve for a variable and then substitute it into the one that will give me the horizontal range.

Answer 5

I was thinking about using y = Voy*t +(0.5)gt^2 to solve for t and then plug that into x = Vox/t. I had another problem similar to this one, but I was not give the velocity in vector form.

Answer 6

We have
v = 8.3 i + 4.9 j ; for t=3
Acceleration = -g j

So we can use v = u + at for vertical direction and get the initial Vy
Vx will remain constant through out the motion.

From here, we can use the usual projectile motion analysis. :)

Answer 7

Okay, I guess that makes sense. I am assuming that u is the velocity found from y = Voy*t +(0.5)gt^2 when y = 0? And then that would allow for the solving of v?

Answer 8

In (v = u + at)
v is the final velocity
u is the initial velocity

We already have v for t = 3 ; we need to find u from the above equation.

Answer 9

So u would be u = 8.3i -24.5j?

Answer 10

a = -g = -9.8
Hence, u = v - at = v + gt
(You probably made a mistake with sign :)

Answer 11

Oh, whoops, I subtracted instead of added. u = 8.3i + 34.3j

Answer 12

So, can you solve the question now ??