A car falls off a ledge and drops to the ground in $0.5 s$ . How high is the ledge from the ground?

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Solution

Step 1: Given quantities:
$u = 0 \frac{m}{s} a = 10 \frac{m}{s^{2}} t = 0.5 s$
Step 2: Formula Used:
$s = u t + \frac{1}{2} a t^{2} v \to f i n a l v e l o c i t y u \to i n i t i a l v e l o c i t y a \to a c c e l e r a t i o n t \to t i m e s \to d i s \tan c e$
Step 3: The calculation for the speed:
Putting the values in the equation, we get
$s = 0 \times 0.5 + \frac{1}{2} \times 10 \times 0.5 \times 0.5 s = \frac{1}{2} \times 10 \times 0.5 \times 0.5 s = 5 \times 0.25 s = 1.25 m$
Hence, the height of the ledge from the ground is $1.25 m$ .

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