Answer:
The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.
Explanation:
The ball has an initial linear kinetic energy and initial rotational kinetic energy which can both be converted into gravitational potential energy. Therefore the hill with friction will let the ball reach higher.
This is because:
If we consider the ball initially at rest on a frictionless surface and a force is exerted through the centre of mass of the ball, it will slide across the surface with no rotation, and thus, there will only be translational motion.
Now, if there is friction and force is again applied to the stationary ball, the frictional force will act in the opposite direction to the force but at the edge of the ball that rests on the ground. This friction generates a torque on the ball which starts the rotation.
Therefore, static friction is infact necessary for a ball to begin rolling.
Now, from the top of the ball, it will move at a speed 2v, while the centre of mass of the ball will move at a speed v and lastly, the bottom edge of the ball will instantaneously be at rest. So as the edge touching the ground is stationary, it experiences no friction.
So friction is necessary for a ball to start rolling but once the rolling condition has been met the ball experiences no friction.
Using
F= mv²/r
4 = 0.5×v² / 2
8 /0.5 = v²
v²=16
v= √16
v= 4 ms-¹
Answer:
The angle is 
Explanation:
From the question we are told that
The distance of the dartboard from the dart is 
The time taken is 
The horizontal component of the speed of the dart is mathematically represented as
where u is the the velocity at dart is lunched
so
substituting values
=> 
From projectile kinematics the time taken by the dart can be mathematically represented as
=> 
=> 
![\theta = tan^{-1} [0.277]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%20%5B0.277%5D)
