Height of the rocket will be <span>h(t)=−<span>12</span>g<span>t2</span>+<span>v0</span>tsinθ+<span>h0</span></span> where
<span>g=9.8<span> m/s2</span></span>
<span><span>v0</span>=86 m/s</span>
<span><span>h0</span>=0 m</span>
<span>θ= angle formed with the vertical
</span>
That's a parabola. You'll solve that for <span>h(<span>tf</span>)=0</span> to find the time of flight.
The horizontal component of the rocket's velocity will be <span><span>vx</span>=<span>v0</span>cosθ</span>. You know that <span>x=<span>vx</span><span>tf</span>=104 m</span> where <span>tf</span> is the time of flight. You can use that relationship to write an expression for <span>tf</span> in terms of <span>v0</span> and θ. Substitute that into the first equation and solve for θ.
Once you've got the parabola figured out, you can easily find the maximum height by finding the vertex, and you've already found the duration of the flight.
Answer:
I hope this a little bit.
Answer:
(a) 1.11sec
(b) 14.37m/s
(c) 31.78m
Explanation:
U = 18m/s, A = 37°, g = 9.8m/s^2
(a) t = UsinA/g = 18sin37°/9.8 = 18×0.6018/9.8 = 1.11sec
(b) Ux = UcosA = 18cos37° = 18×0.7986 = 14.37m/s
(c) R = U^2sin2A/g = 18^2sin2(37°)/9.8 = 324sin74°/9.8 = 324×0.9613/9.8 = 31.78m
Answer:
Option D. is correct.
Explanation:
The object's mechanical energy refers to the sum of the potential and kinetic energies of the object. When an object falls, its potential energy (PE) decreases, and its kinetic energy (KE) increases. The increase in kinetic energy is exactly equal to the decrease in potential energy.
Option D. is correct.
I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. "<span>The charge can be located anywhere, since flux does not depend on the position of the charge as long as it is inside the sphere" gives the answer to the question. I hope that this answer has come to your help.</span>
