Answer:
7.68m/s
Explanation:
At point A, height h= 0m where v=?
At point B, height of h = 3.01 m above the top of the ramp v= 0 m/s
The conservation of energy here will be energy at Point A equals energy at point B
g= acceleration due to gravity= 9.81m/s^2
(mgh + 1/2 mv^2)A = (mgh + 1/2 mv^2)B
✓But height h at point A= 0 m then energy at point A= 1/2 mv^2
✓At point B , the velocity v= 0 m/s, then Energy at point B= mgh
✓ if we equate both energy at point A and B we have
mgh=1/2 mv^2
Let us make v the subject of formula, and cancel out the "m"
v=√2gh
Then substitute the values
V= √2×9.81×3.01
V=7.68m/s
Hence, his initial speed (in m/s) at point A is 7.68m/s
