A 0.145-kg baseball pops straight up from the ground at 21.4 m/s(a) What maximum height does it reach? (b) Find the work done by gravity on the baseball during its upward flight. (c) Find the work done by gravity for the entire round trip, from launch to when the ball hits the ground.

(a) h = 23.34 m

(b) W = - 33.2 J

(c) W = 0 J

Explanation:

(a)

We will use the third equation of motion to find out the height reached by the ball:

where,

g = acceleration due to gravity = - 9.81 m/s² (-ve sign due to upward motion)

h = height reached by ball = ?

vf = final speed = 0 m/s

vi = initial speed = 21.4 m/s

Therefore,

h = 23.34 m

(b)

The work done by gravity on the ball will be equal to its potential energy:

W = mgh

where,

W = Work done = ?

m = mass of ball = 0.145 kg

Therefore,

W = (0.145 kg)(- 9.81 m/s²)(23.34 m)

W = - 33.2 J

(c)

The total work done will be the sum of the work done during upward and downward motion. Here the magnitude of the work done during upward motion will be the same as the magnitude of the work done during downward motion with opposite direction. Therefore,

W = -33.2 J + 33.2 J

W = 0 J

we need to know the time interval during which the force acts. impulse of force = momentum change

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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