A ball of mass 100 g is thrown vertically upward at a speed of 25 ms⁻¹. If no energy is lost, determine the height it would reach. If the ball only rises to 25m, calculate the work done against air resistance. Also, calculate the force of friction. (31.9 m, 6.7 J, 0.3 N) 



Given:

Mass of the ball = m = 100 g = 0.1 kg (conversion SI unit)
Speed of the ball = v = 25 m s⁻¹
Value of g = 9.81 m s⁻²

To Find:

(a) Height of the ball reached = h = ?
(b) Work done against the air resistance at height (h = 25) = W = ?
(c) Force of friction = F = ?

Solution:

(a) Height of the ball reached = h = ?

According to the law of conservation of energy

Gain in K.E = Loss in P.E.

`\ frac{1}{2}`mv² = mgh

h = `\ frac{v²}{2g}`

by putting values

h = `\ frac {(25 m s⁻¹)²}{ 2 ｘ 9.81 m s⁻² }`

h = `\ frac {625 m² s⁻²}{ 19.62 m s⁻² }`

h = 31.855 m 

or

h = 31.9 m  ----------------Ans


(b) Work done against the air resistance at height (h = 25) = W = ?

When the ball rises only 25 m then according to the law of consecration of energy

Loss in K.E. = Gain in P.E. + Work done against air resistance

or

Work done against air resistance = Loss in K.E. = Gain in P.E.

W = `\ frac{1}{2}`mv² - mgh 

putting values

W = `\ frac{1}{2}` ｘ 0.1 kg (25 m s⁻¹)² - 0.1 kg ｘ 9.81 m s⁻² ｘ 25 m

 
W = 31.25 kg m² s⁻² - 24.525 kg m² s⁻² 

W = 6.725 kg m² s⁻² 

Or

W = 6.725 J ------------Ans


(c) Force of friction = F = ?

As we know that W = F . h so, 

F = `\frac {W}{h}`

by putting values

F = `\frac {6.725 J }{25 m}`

F = 0.269 N

or

F = 0.3 N -------------Ans.


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