To solve this problem, we can use the equation of motion:
v^2 = u^2 + 2as
Where:
v is the final velocity (which is 0 m/s when the ball comes to a stop)
u is the initial velocity (50 m/s)
a is the acceleration (which we can assume to be constant for simplicity)
s is the distance traveled up the incline (80m)
Plugging in the values we have:
0 = (50)^2 + 2a(80)
0 = 2500 + 160a
-2500 = 160a
a = -2500 / 160
a = -15.625 m/s^2
Now we can find the velocity of the ball 2.5s after starting up the incline:
v = u + at
v = 50 + (-15.625)(2.5)
v = 50 - 39.0625
v = 10.9375 m/s
Therefore, the velocity of the ball 2.5s after starting up the incline is approximately 10.9 m/s.
A ball is travelling at 50 m/s begins to roll up an inclined plane before coming to the Ball comes to a stop 80m up the incline what was the velocity of the ball 2.5s after starting up the incline The answer is 10.9
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