Asked by Anonymous
In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that in the figure below and has a range of 15 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
Answers
Answered by
bobpursley
range will be proportional to speed,and time in air.
so how does speed affect time in air?
consider time in air...
hf=h0 + vosinTheta*time-1/2 g t^2
0=0+ t(V-1/2 g t)
so t=2V/g so time in air is directly proportinal to initial speed.
range is then proportional to v^2
so if it was 15m, it is now 15*4 since speed was doubled
so how does speed affect time in air?
consider time in air...
hf=h0 + vosinTheta*time-1/2 g t^2
0=0+ t(V-1/2 g t)
so t=2V/g so time in air is directly proportinal to initial speed.
range is then proportional to v^2
so if it was 15m, it is now 15*4 since speed was doubled
Answered by
Anonymous
In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a trajectory similar to that shown in the figure below and has a range of 17 m. Suppose the launch speed is doubled, and the projectile is fired at the same angle above the ground. What is the new range?
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