Two stones are thrown vertically upwards from the same point with the same velocity of 20m/s but at an interval of 2s. When they meet, the second stone is rising at 10m/s.
a) For what time is the second stone in the air before they meet?
b) What is the velocity of the first stone when they meet?
c) State any assumptions you make in obtaining your answers.
Help is greatly appreciated. Thanks.
For A. Doesn't that just involve how long it takes for gravity (9.81 m/s) to halve the velocity of the stone? And then you can use that answer for B.
So basically a force of 9.81 m/s (idk in newtons) acting against something with a force of 20 m/s untiil the latter reaches 10 m/s.
You have a deceleration on both of 9.82m/s^2, you could just calculate how long it takes for gravity to decelerate stone 2 down to 10m/s and you have the answer to A. Once you have the time for the meeting point, calculate how much gravity will have decelerated stone 1 during that time (don't forget to add 2 seconds since stone B was thrown later).
We actually had this exact same question in our last physics test :v:
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