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A circus acrobat is launched from a cannon that is aimed at 75∘ above the ground. He leaves the barrel at a height of 3.5 meters, travelling at 7 meters per second. If the net must be placed at a height of 5.5 meters above the ground for safety, at what horizontal distance from the end of the cannon barrel should the centre of the net be placed so as to catch the acrobat exactly at that position? (Ignore air resistance, assume the acrobat's position can be approximated by a point and that g=10m/s2.

The centre of the net should be located ? meters from the end of the cannon barrel.

Guest May 14, 2017
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The equation for the vertical height of a body moving under gravity is:

 

s = ut + Β½at2  

 

Our acrobat starts higher than 0 metres, so I'll modify the equation so it shows him starting at 3.5m,

s = 3.5 + ut + Β½at2

 

Solve this quadratic to find t when s = 5.5 to determine how long it takes him to reach the 5.5m heights. One of the times is that for the acrobat ascending, the other is for his descending into the net.

 

Once you know the time he hits the net, you can determine how far the horizontal component of velocity has taken him.

 

πŸŽͺ  πŸŽ  🎠 🎠 🎭 

Badinage  May 14, 2017

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