(Thought question) At noon On June 21 will the shadow length of 42 degrees north latitude be shorter or longer than it is on Mar
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The person's horizontal position is given by
and the time it takes for him to travel 56.6 m is
so your first computed time is the correct one.
The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity at time
is
which tells us that he would reach the ground at about . In this time, he would have traveled
But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity would have been a bit smaller than
.
Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.