A skateboarder starts at point A on the ramp and rises to point B, a maximum height of h = 3.01 m above the top of the ramp. If the amount of work done against friction is insignificant, determine his initial speed (in m/s) at point A

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

are there answer choices to this question? this seems like a question where you have options to choose from.