so, how far have you gotten? We know that the horizontal distance at time t is
d = v cosθ t
So, for this problem, if we let
θ = angle of 1st shot
Ø = angle of 2nd shot
k = delay between shots
t = flight time of 1st shot
1000 cosθ t = 1000 cosØ (t-k)
cosθ/cosØ = (t-k)/t
and we know that the vertical speed of the 1st shot is v(t) = 1000 sinθ - 10t
so that it spends t=200sinθ seconds in the air.
So, now we have
cosθ/cosØ = 1 - k/(200sinθ)
So, you need to find the max value of t (or sinθ, since t=200sinθ) which satisfies these conditions.
The illustration for this problem can be seen at screenshots(dot)firefox(dot)com/atwXACAA9akaoAji/null
The army is testing out a new prototype artillery cannon with an uncommonly high muzzle velocity of 1000 m/sec . The design bugs haven't been fully worked out yet, so the cannon has to wait at least a full minute between shots (reloading, etc.). The cannon can be aimed with an angle between 0 and 90 degrees with respect to the horizontal.
A particularly demanding exercise is being conducted, in which the cannon fires an initial shot into the air, and while the first shell is still in the air, a second shot is fired (at a different angle) so that both shells impact the target (at ground level) simultaneously.
What is the maximum horizontal distance from the cannon to the target, such that this is possible?
Details and Assumptions:
Assume level ground, with no air resistance.
The gravitational acceleration is 10 m/s².
Give your answer in meters, to the nearest whole meter.
For the sake of this problem, ignore the Earth's curvature.
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