# Homework Help: Calculating the Vertical component of the Velocity

1. Oct 25, 2014

### bronteogrady

1. The problem statement, all variables and given/known data
A physics class researching projectile motion constructs
a device that can launch a cricket ball. The launching
device is designed so that the ball can be launched at
ground level with an initial velocity of 28.0 m s–1 at an
angle of 30.0 degrees to the horizontal.

2. Relevant equations
Calculate the vertical component of the velocity of the ball:
a initially
b after 1.00 s
c after 2.00 s.
3. The attempt at a solution
Done A how do I do B and C??

Thankyou

2. Oct 25, 2014

### Orodruin

Staff Emeritus
Hi and welcome to Physics Forums!

We require you to show some effort in working towards a solution before we provide help and guidance. What are your own thoughts on how to approach B and C?

3. Oct 25, 2014

### bronteogrady

I thought about using the acceleration equations though I have not used these before with a degrees in the mix and was unsure if I needed to include this.

4. Oct 25, 2014

### Orodruin

Staff Emeritus
So how do you think the ball is accelerated? Which component of the velocity is affected?

5. Oct 25, 2014

### bronteogrady

The vertical component because horizontal does not use the acceleration equations

6. Oct 25, 2014

### Orodruin

Staff Emeritus
Yes, so try applying that and show us what you get.

7. Oct 25, 2014

### bronteogrady

so using the initial vertical component at the initial velocity in the equation v=u+at?

8. Oct 25, 2014

### Orodruin

Staff Emeritus
Yes. As you said yourself, this is the only component of velocity that is influenced by the acceleration.

9. Oct 25, 2014

### bronteogrady

Thankyou!!! Now I have it is seems easy! just wondering, why is the 9.8 ms-2 negative? is is because when it is moving up it is going in the opposite direction? If so does this mean that when is is moving downwards the 9.8ms-2 would be positive?

10. Oct 25, 2014

### Orodruin

Staff Emeritus
This depends completely on how you have defined your coordinates. If you have defined the up direction as the positive y-direction, then the gravitational acceleration will be in the negative y-direction, i.e., down. If you did it the other way around and defined the down direction as positive, then you would have a positive gravitational acceleration, but the sign of the initial velocity in that direction would be changed as well.

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