The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by $v\sin39^\circ$ , where $v$ here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving

$\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(v\sin39^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm m)$

$\implies v=11\,\dfrac{\mathrm m}{\mathrm s}$

6 0

2 years ago

A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the

crimeas [40]

Answer:

Explanation:

The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²

where Ф is the angle of inclination, R is the radius, k is the radius of gyration.

The potential energy of the system is given as ; PE = mgh

The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.

The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2

Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

7 0

2 years ago