Using the concepts of motion, we got that 0.816m is the maximum vertical height the box will reach if its initial velocity at the bottom of the ramp is 4.0 m/s.
We use Newton`s third equation of motion,
v² -u²=2aS---------(eq1)
where
v is the final velocity, and u is the initial velocity and a is the acceleration of the object and S is the displacement of the object.
We know, that at the maximum height, final velocity will be =0
Acceleration will be acceleration due to gravity(g)=9.8m/sec² .
Putting the values in (eq1)
=>0-(4)² = 2×(-9.8)×S
=>2×9.8×S = 16
=>S = [(16)/(2×9.8)]
=>S = 8/9.8
=>S =0.816m
Hence, the the maximum vertical height the box will reach if its initial velocity at the bottom of the ramp is 4.0 m/s is 0.816m
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