A shot putter throws the 726 kg steel ball a horizontal...

A shot putter throws the 7.26-kg steel ball a horizontal distance of 14.7 m. The shot was released at a height of 2.71 m and the initial velocity made and angle of exactly 45 degrees with horizontal. a) What was the initial speed of the ball in m/s? v = b) If the ball moved 1.82 m (in a straight line) while it was being thrown, then what was the magnitude of the net force, in N, on the ball as it was being thrown? Fnet =

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Transcript

00:01 Here we are given a projectile motion.

00:08 The vertical distance is given, that is x is equal to 14 .7 meter.

00:18 Horizontal distance y is equal to 2 .71 meter and here theta is equal to 45 degree.

00:28 Theta represents the angle of launch.

00:32 So here, let us see the shape of the projectile.

00:42 Here, this is the shape of the projectile mentioned here.

00:48 There is an angle theta here with the horizontal.

00:57 This is angle theta.

00:59 This is horizontal which is represented as v0 cos theta.

01:03 And here the vertical component is represented as v0 sine theta.

01:08 This is v0 and here this represents the value of y which is 2 .71 meter and this represents the value of x.

01:22 Now let us move on to the first part a part here.

01:29 First we need to calculate the velocity in horizontal direction.

01:38 So here the velocity in horizontal direction is given by v0 x is equal to v0 cos theta which is equal to v0 cos 45 or this is equal to v0 by root 2 let this be equation number 1 next we need to calculate the velocity in vertical direction velocity in vertical direction is given by v0 y is equal to v0.

02:12 Y is equal to v0 sine theta this is equal to v0 sine 45.

02:19 This turns out to be v0 by root 2.

02:23 Let this be equation number 2...

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