A rocket starts vertically upward with speed `v_(0)`. Show that its speed `v` at height `h` is given by `v_(0)^(2)-v^(2)=(2hg)/(1+h/R)` where `R` is the radius of the earth and `g` is acceleration due to gravity at earth's suface. Deduce an expression for maximum height reachhed by a rocket fired with speed `0.9` times the escape velocity.
Text Solution
AI Generated Solution
To solve the problem, we will use the principle of conservation of mechanical energy, which states that the total mechanical energy (kinetic energy + potential energy) of the rocket remains constant if we neglect air resistance.
### Step-by-Step Solution:
1. **Identify Initial and Final States:**
- The rocket starts from the surface of the Earth with an initial speed \( v_0 \) and at a height \( h = 0 \).
- At height \( h \), the rocket has speed \( v \).
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