Vertical Projectile Motion Problem

[TheRedDevil18]

Homework Statement

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?

Homework Equations

equations of motion

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

Could someone please check?

 

[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.

 

[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.

 

[CWatters]

For 3)

v = u + at

t = (v - u)/a

v=-25

t = (-25 - 30)/-9.8

= 5.6 seconds

 

[CWatters]

So all looks about right.

 

[TheRedDevil18]

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.

Why did you use t=8?, because wouldn't that include the height that it was thrown up at?, they wanted the height of the building

 

[haruspex]

Why did you use t=8?, because wouldn't that include the height that it was thrown up at?, they wanted the height of the building

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.

 

[TheRedDevil18]

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.

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 wouldn't you have to use 8-6= 2's for the time?, sorry just a little confused

 

[haruspex]

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 wouldn't you have to use 8-6= 2's for the time?, sorry just a little confused

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.

 

[TheRedDevil18]

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.

Think that cleared things up, thanks

 

[CWatters]

Why did you use t=8?, because wouldn't that include the height that it was thrown up at?, they wanted the height of the building

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 separate 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.