t = 0.50 s
"v_{0}=12 m\/s"
This is a question on projectile motion.
The distance from the cliff's edge after truck leaves is the range.
"range=v_{0}t"
Where "v_{0}" = initial velocity
t = time
Range = 0.5 × 12 = 6 m
For the vertical components :
We will use the following equation of motion :
"v=v_{0}+at"
a = acceleration due to gravity
"v" = final velocity
"v=12+5=17 m\/s"
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