**Problem 6:** A ball of mass 100 grams 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 25 m, calculate the work done against the air resistance. Also calculate the force of friction.

**Solution**

Explanation: The problem has 3 parts;

- To determine the height of the ball neglecting air resistance.
- In the presence of air resistance, the ball goes to a height of 25 m. Here we have to calculate the work done against the air resistance.
- Friction with air (air resistance) exerts a force on the ball. We have to calculate that force.

I: **Given** Initial velocity, v_{i} = 25 ms^{-1}, Final velocity, v_{f} = 0 ms^{-1}

Acceleration,
g = – 9.8 ms^{-2}

**Find ** Distance, s = h.

**Formula** 2as = v_{f}^{2} – v_{i}^{2}

Put the values,

2(-9.8)h = 0^{2} – (25)^{2} OR -19.6h = -625

OR h = 625/19.6 = 31.88 = 31.9 m

Therefore, the ball goes 31.9 m high in the air.

**II: Given** v_{i} = 25 m/s v_{f} = 0 m/s

s = h = 25 m

**Find** Air resistance, f

Here, when the ball is thrown up, it has some K.E which converts to P.E and work against the air resistance. According to the law of conservation of energy, the K.E,at any instant of time, will be equal to the gain in P.E plus work done against the air resistance.

K.E = Gain in P.E + Work done against the air resistance. … (A)

Now, K.E = ½ mv^{2} = ½ × (100 × 10^{-3}) ×
25 × 25 = 31.2
J … (1)

(100 × 10^{-3} = 0.1) is the mass of the ball
converted to kilograms.

P.E = mgh = 0.1 × 9.8 × 25 = 24.5 J … (2)

Put values from (1) and (2) in (A)

31.2 = 24.5 + Work done against air resistance

∴ Work done against air resistance = 31.2 – 24.5 = 6.7 J

**III:** Now work done = W = Fd, or F = W/d. Put values of W and d to calculate the force due to air resistance,

F = 6.7/25 = 0.268 = 0.3 N

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