A cyclist has a final velocity of 30 m/s after coasting down hill for 2 seconds. If the cyclist underwent a constant acceleration of 7.5m/s2 during the descent, what was his initial velocity at the top of the hill?

To find the initial velocity of the cyclist at the top of the hill, we can use the following equation of motion:

v = u + at

Where:
v = final velocity = 30 m/s
u = initial velocity (to be found)
a = constant acceleration = 7.5 m/s^2
t = time = 2 seconds

Rearranging the equation, we have:

u = v - at

Substituting the given values:

u = 30 m/s - (7.5 m/s^2)(2 s)
u = 30 m/s - 15 m/s
u = 15 m/s

Therefore, the initial velocity of the cyclist at the top of the hill was 15 m/s.

To find the initial velocity of the cyclist at the top of the hill, we can use the equation of motion:

v = u + at

Where:
v = final velocity = 30 m/s
u = initial velocity (unknown)
a = acceleration = 7.5 m/s^2
t = time = 2 s

Rearranging the equation, we have:

u = v - at

Substituting the given values, we get:

u = 30 m/s - (7.5 m/s^2)(2 s)
u = 30 m/s - 15 m/s
u = 15 m/s

Therefore, the initial velocity of the cyclist at the top of the hill was 15 m/s.