- neon signs
- fireworks
- rockets during the ascent
- meteors
- fireflies
- stars
The kinematic equations of motion that apply here are<span>y(t)=votsin(θ)−12gt2</span>and<span>x(t)=votcos(θ)</span>Setting y(t)=0 yields <span>0=votsin(θ)−12gt2</span>. If we solve for t, we obtain, by factoring,<span>t=<span>2vsin(θ)g</span></span>Substitute this into our equation for x(t). This yields<span>x(t)=<span><span>2v2cos(θ)sin(θ)</span>g</span></span><span>This is equal to x=<span><span>v^2sin(2θ)</span>g</span></span>Hence the angles that have identical projectiles are have the same range via substitution in the last equation is C. <span> 60.23°, 29.77° </span>
Answer:
The acceleration that the jet liner that must have is 2.241 meters per square second.
Explanation:
Let suppose that the jet liner accelerates uniformly. From statement we know the initial (
) and final speeds (
), measured in meters per second, of the aircraft and likewise the runway length (
), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed (
), measured in meters per square second:
If we know that 
, 
and 
, then the acceleration that the jet must have is:
The acceleration that the jet liner that must have is 2.241 meters per square second.
