Consider a particle launched at a horizontal velocity v_0 from a height h above

Question

Consider a particle launched at a horizontal velocity v_0 from a height h above the ground. Derive an expression for the time it takes the projectile to strike the ground. Ignore air resistance. Derive an expression for the maximum horizontal distance ("range") travelled by the projectile. Ignore air resistance. Let's include the effect of constant air resistance on the projectile: the acceleration of the particle is now given by a^rightarrow = - epsilon gi - gj, where g - 9.8 m/s^2 and epsilon is a very small non-negative constant (epsilon

Explanation / Answer

1) if we ignore the air resistance then we have to consider two components of the velocity V0. The x component of V0

is Vx= Costheta where theta is the angle made with base.

the Y component of velocity Vy = V0 Sin theta-gt

Using the V=u+ at equation and taking a=g we get total taken time T = 2V0Sintheta/g

2) We use V^2=U^2 +2aS equation to find the height expression

Here we consider H=S

and initial velocity V0, and a=g

so using them in the equation we get H=V0^2Sin^2theta/2g

3. now we consider air resistance that means a=-egi-gj , using the same total time taken T= 2V0Sintheta/-egi-gj

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