Question

a stone is thrown vertically up from the top of a cliff with a velocity v. air resistance is negligible. what is the variation with time of the velocity of the stone until it hits the ground?
options: a, b, c, d (with corresponding velocity - time graphs)

Explanation:

Step1: Analyze Motion Type

The stone is thrown vertically up, so it undergoes uniformly accelerated motion (free - fall with acceleration \(g\) downward). The velocity - time relationship for uniformly accelerated motion is \(v = v_0+at\), where \(a=-g\) (taking upward as positive).

Step2: Analyze Velocity Change

• Initially, the velocity \(v_0\) is positive (upward). As time \(t\) increases, the velocity decreases linearly (because \(a = - g\) is constant) until it reaches zero at the maximum height.
• After reaching the maximum height, the stone starts to fall down. Now, the velocity becomes negative (downward) and its magnitude increases linearly with time (since the acceleration is still \(g\) downward, so \(a=-g\), and for the downward motion, the initial velocity for this phase is 0, and \(v = 0+(-g)t\) where \(t\) is the time elapsed since reaching the maximum height).
• So the velocity - time graph should be a straight line with a negative slope (because acceleration is constant and negative) that crosses the time axis (when velocity is zero at maximum height) and then goes negative with increasing magnitude. Looking at the options, option A has a straight line with a constant slope (negative, since it goes from positive initial velocity to negative velocity as time increases), which matches the motion of the stone (uniformly accelerated motion with constant acceleration \( - g\)). Option B has a kink, which is wrong as acceleration is constant. Option C is a curve, which is for non - uniform acceleration. Option D has a non - linear change in slope, which is also wrong.

Answer:

A