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
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.