A tennis ball, starting from rest, rolls down the hill in...

Name: A tennis ball, starting from rest, rolls down the hill in the drawing (h = 1.54 m). At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin-walled spherical shell, and determine the range x. (g = 9.80 m/s²)
Uploaded: 2021-07-07T06:39:32-08:00
Duration: 3 min 53 s
Channel: Darshan Maheshwari
Description: A tennis ball, starting from rest, rolls down the hill in the drawing (h = 1.54 m). At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin-walled spherical shell, and determine the range x. (g = 9.80 m/s²)

A tennis ball, starting from rest, rolls down the hill in the drawing (h = 1.54 m). At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin-walled spherical shell, and determine the range x. (g = 9.80 m/s²)

Transcript

00:01 Talk about this question we are given that a tennis ball starting from rest it starts from rest from here it rolls down rolls down the hill in the drawing and the height is given us 1 .54 meters at the end of the hill the ball becomes airborne so after going down going up it becomes airborne so that it leaves the ground and it moves as a project time motion the angle at which it leaves is 35 degrees we need to find the range that is the x the value of x in this project type motion and we have to treat the ball as a thin wall spherical shell all right so the first agenda should be to find the velocity with which it leaves from here let's call this v0 because that only we can find the range because range is dependent on the velocity at which it is leaving the project and the range is given by v0 square sine 2 theta or g so v0 is something which is unknown theta is already given as 35 degrees so agentized to find v0 now what we can do is i can conserve the energy so let's call this point a let's use the different color over here let's call this point a let's call this point b so you're gonna conserve the energy we're gonna do energy conservation energy conservation at point a and b so at point a we just have the potential energy because it is released from the rest the potential energy at a is going to be mgh.

01:37 And then we talk about the potential, the energy at b.

01:41 So at b, the height is zero.

01:42 So there is no potential energy, but there is just angular kind of rotational kinetic energy and linear kinetic energy.

01:51 So rotational kinetic energy is given by i omega square over 2.

01:55 And this is given by half mv not square.

02:00 So m gh remains as it is i of a cylindrical shield is given by 2 over 3 m r square and 1 over 2 omega square remains as it is plus 1 over 2 m v0 square so m is cancelled we are left with gh is equal to this 2 is also cancelled 1 over 3 omega and r are written together and this is 1 half v0 square since it is rolling there is a relation between the velocity and the angular velocity that v0 is equal to omega not r.

02:31 We're going to use that and replace omega -r by v -0...

Question

A tennis ball, starting from rest, rolls down the hill in the drawing (h = 1.54 m). At the end of the hill the ball becomes airborne, leaving at an angle of 35° with respect to the ground. Treat the ball as a thin-walled spherical shell, and determine the range x. (g = 9.80 m/s²)

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