A car falls off a ledge and drops to the ground in $0.5 s (g = 10 m s^{2})$ . calculate:

How high is the ledge from the ground?
What is its speed on striking the ground?
What is its average speed during 0.5 s?

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Solution

Time of free fall,

$s = u t + \frac{1}{2} a t^{2}$
initial velocity is 0 so

$h = \frac{1}{2} g t^{2} = \frac{1}{2} 10 \times {0.5}^{2} = 1.25 m$
(c) height from ground is $1.25 m$

Apply kinematic equation of motion

$v^{2} - u^{2} = 2 a s$

$v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 1.25} = 5 m s^{- 1}$

(a)Speed of car striking the ground is $5 m s^{- 1}$
(b)Average speed, $v_{a v g} = \frac{v - 0}{h} = \frac{5}{1.25} = 4 m s^{- 1}$

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