Free Fall Physics Example 4

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Example 4

medium
A ball is thrown straight up with an initial velocity of 20 m/s20 \text{ m/s}. Using g=10 m/s2g = 10 \text{ m/s}^2: (a) How long until it reaches its highest point? (b) What maximum height does it reach? (c) What is its velocity when it returns to the starting height?

Solution

  1. 1
    (a) At the top, v=0v = 0. Using v=ugtv = u - gt: 0=2010t0 = 20 - 10t, so t=2 st = 2 \text{ s}.
  2. 2
    (b) h=ut12gt2=20(2)12(10)(4)=4020=20 mh = ut - \frac{1}{2}gt^2 = 20(2) - \frac{1}{2}(10)(4) = 40 - 20 = 20 \text{ m}.
  3. 3
    (c) By symmetry, it returns at the same speed but downward: v=20 m/sv = -20 \text{ m/s} (or 20 m/s20 \text{ m/s} downward).

Answer

(a)  2 s;(b)  20 m;(c)  20 m/s downward(a)\; 2 \text{ s}; \quad (b)\; 20 \text{ m}; \quad (c)\; 20 \text{ m/s downward}
In free fall (ignoring air resistance), an object thrown upward decelerates at gg, stops momentarily at the peak, then accelerates back down. The motion is symmetric — same speed at a given height going up and coming down.

About Free Fall

Motion under gravity alone, with no air resistance — all objects in free fall accelerate at g9.81g \approx 9.81 m/s² regardless of mass.

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