ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: sajib_mostofa on June 30, 2010, 06:08:17 pm
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Anyone care to show me how to do this Q?
A balloon is moving vertically upwards with a uniform spee of 20m/s. The balloon drops an object when it is 300m from the ground.
a) How long does it take to reach the ground? What distance does it travel in doing so?
b) What is the height of the balloon above the ground when the objects is released?
Answers
a) 10 s; 340m
b) 500m
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lol I was just doing this question yesterday.
a): using s=ut+(1/2)at^2, given that the object has to fall 300m so s=300, and the initial velocity (u) is -20 m/s, and the acceleration due to gravity is 10, sub in those values and you will get 10 seconds for t.
However, the distance it travels is slightly further, because the object continues to rise for a short while after it it released by the balloon. To find how much, you have to find t when v=0 and sub that into the equation for s, and you will find it actually rises a further 20m, so the total distance travelled is 340m.
b): since it takes 10s to hit the ground, and the balloon is rising at 20 m/s and started at 300m when the object was released, after at 10s it will be at 500m.
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Cheers man :)