Knight riders physics thread

[knightrider]

draw a velocity-time graph

How would you do that?

how would you do this question

Two physics students conduct the following experiment from a very high bridge.Thao drops a 1.5 kg shot-put from a vertical height of 60.0 m while at exactly the same time Benjamin throws a 100 g mass with an initial downwards velocity of 10.0 m s^−1 from a point 10.0 m above Thao.

At what time will the 100 g mass overtake the shot-put?

I think you have to solve equations simultaneously but what equations do you use and how do you get them

can anyone help

[knightrider]

Hi guys how would you do this question?
 
I have worked out that the shot-put takes 3.5 seconds to reach the ground
I have worked out the 100 g mass takes 2.9 seconds to reach the ground.

Two physics students conduct the following experiment from a very high bridge.bridge. Thao drops a 1.5 kg shot-put from a vertical height of 60.0 m while at exactly the same time Benjamin throws a 100 g mass with an initial downwards velocity of 10.0 m/s from a point 10.0m above Thao.

At what time will the 100 g mass overtake the shot-put?

[knightrider]

bump

[dankfrank420]

Is the answer 1 second? (Assume g = 10)

Use the formula x = ut +at^2

You want to find where X thao = X Benjamin, rewritten as X thao = X Thao - 10 (as Benjamin starts from 10 m higher)

X thao = 0t + .5(10)^2
X Thao + 10 = -10t + .5(10)^2

X thao = X thao - 10
.5(10)^2 = 10t + .5(10)^2 - 10
---> t = 1

[knightrider]

Is the answer 1 second? (Assume g = 10)

Use the formula x = ut +at^2

You want to find where X thao = X Benjamin, rewritten as X thao = X Thao - 10 (as Benjamin starts from 10 m higher)

X thao = 0t + .5(10)^2
X Thao + 10 = -10t + .5(10)^2

X thao = X thao - 10



.5(10)^2 = 10t + .5(10)^2 - 10
---> t = 1

Thanks dankfrank420

[knightrider]

How would you add the following forces?

60 N east and 80 N south

[Zlatan]

You can represent those forces through vectors. So you can construct a right angle triangle with those forces.

After doing that, you have to find the length of the hypotenuse (pythagoras) which would give you the magnitude of the resultant force.

To find the direction, just find the angle to the horizontal using trig ratios.

Hope this helped 

[Zlatan]

I forgot to mention this, after finding the angle you can leave it as true bearings or compass bearings. Personally, I think compass bearing are better because the direction that are given in the question are compass directions as well.