Question

A ball is rolling at 4.80m/s over level ground when it encounters a ramp, which gives it an acceleration of -0.875m/s^2. If the ramp is 0.750m long, what is the final velocity of the ball when it reaches the top of the ramp?

Answer 1

4.66 m/s The distance the ball will move when under constant acceleration will be d = VT + 0.5A T^2 where d = distance traveled V = initial velocity A = acceleration T = Time Substituting the known values gives. 0.750 m = 4.80 m/s * T - 0.5 * 0.875 m/s^2 * T^2 0.750 m = 4.80 m/s * T - 0.4375 m/s^2 * T^2 0 m = 4.80 m/s * T - 0.4375 m/s^2 * T^2 - 0.750 m We now have a quadratic equation with a = -0.4375, b = 4.80, and c = -0.750. Using the quadratic formula, we find the roots 0.158540972 seconds and 10.8128876 seconds. Then 10+ second root would happen if the ramp extended further and the ball rolled up until it stopped, then started rolling backwards. Since the ramp isn't that long, the 0.1585 second is the root we desire. And the velocity will be the initial velocity, minus the deceleration caused by climbing the ramp. And that would be the acceleration multiplied by the time spent accelerating. So 4.80 m/s - 0.875 m/s^2 * 0.1585 s = 4.80 m/s - 0.13869 m/s = 4.66 m/s

Answer 2

Given that the ball was rolling with an initial velocity of 4.80 m/s, when it encountered the ramp.

Given that the accerelation with which it rolled over the ramp was
`-0.875 m/s^2` and that the ranp is 0.750 m long.

The final velocity of an object with an initial velocity, u, with an accerelation, a, moving through a distance, s is given by


`v^2=u^2+2as`

Thus, the final velocity of the ball when it reaches the top of the ramp is given by


`v^2=(4.80)^2+2(-0.875)(0.750) \\ \\ =23.04-1.3125=21.7275 \\ \\ \Rightarrow v=√(21.7275)=\bold{4.66 \ m/s}`