# Homework Help: Vertical Projectile Motion Problem

1. May 11, 2013

### TheRedDevil18

1. The problem statement, all variables and given/known data

A ball is projected vertically upward with a velocity of 30m/s. It strikes the ground after 8's.

1.1) Determine the height of the building
1.2) Find the height of the ball relative to the ground as well as its velocity at t=3 and t=5
1.3) At what time will its velocity be 25m/s, downward?

2. Relevant equations

equations of motion

3. The attempt at a solution

1.1) Okay, I worked out the total time to go up and got 3's and then doubled it for coming down and got 6's (edge of the building). 8-6=2's coming down from top of the building.
delta y=vi*t+0.5*a*t^2
height of building = 79.6m

1.2) at t=3's, v=0
delta y = 45.9m
Relative to ground:
79.6+45.9=125.5m

at t=5's
vf = vi+at
= 0+(9.8)*5
= 49m/s - 30m/s
= 19m/s

delta y = 19.6m+79.6
= 99.2m relative to the ground

1.3) vf = vi+at
t = 2.55+3
= 5.55's

2. May 11, 2013

### CWatters

I assume it was thrown from the top of the building..

For 1) I used

S = ut + 0.5at^2

= 30*8 + 0.5(-9.8)82
= 240 - 313.6
= -73.6m

If you assume g = 10m/s/s then you get

= 240 - 320
= - 80m

-ve because it lands that distance below the height at which it was thrown from.

3. May 11, 2013

### CWatters

For 2)
t=3
S = ut + 0.5at^2
S = 30*3 + 0.5(-9.8)32
S = 90-44.1
S = 45.9m (above the top of the building)

Height above ground = 45.9 + 73.6 = 119.5m

Repeat with t=5

Edit: Gives 27.5 + 73.6 = 101.1m above ground.

Last edited: May 11, 2013
4. May 11, 2013

### CWatters

For 3)

v = u + at

t = (v - u)/a

v=-25

t = (-25 - 30)/-9.8

= 5.6 seconds

5. May 11, 2013

### CWatters

6. May 11, 2013

### TheRedDevil18

Why did you use t=8?, because wouldnt that include the height that it was thrown up at?, they wanted the height of the building

7. May 11, 2013

### haruspex

The equation used tells you the height relative to the starting point after t seconds. It doesn't matter what route it takes to get there. E.g. note that t=0 and t=60/9.8 both give s=0.

8. May 12, 2013

### TheRedDevil18

It took 3 seconds to reach max height after it was thrown and another 3 seconds to come down to its starting point(edge of building) so wouldnt you have to use 8-6= 2's for the time?, sorry just a little confused

9. May 12, 2013

### haruspex

Yes, but it's moving the opposite direction then. You seem to want to break it into stages, but there's no need. You can treat it as one smooth movement - in 8 seconds it will rise, then fall, and reach a point 73.6m below the start point. The quadratic equation encapsulates all of that.

10. May 12, 2013

### TheRedDevil18

Think that cleared things up, thanks

11. May 12, 2013

### CWatters

Looks like you already got it but..

t=0 is "a" time that it is it is at the top of the building
t=8 is the time it is at the bottom of the building (eg the ground)

so the displacement S is the height of the building.

As haruspex said. The equation works for any point on the path even when that's below the starting point.

On a seperate but related issue.. It's worth remembering that some problems have two possible solutions. For example a ball thrown up from the ground at an angle so that it lands on the roof of a building would have two solutions for the time of flight to height h. One on the way up and the other on the way down.