A swimmer’s speed along the river (downstream) is 20 Km/h and can swim upstream at 8 km/h. calculate the velocity of the stream and the swimmer’s possible speed in still water. Solution Swimmer’s speed downstream = 20 km/h Swimmer’s speed upstream = 8 km/h Now, velocity of stream =? Also, swimmer’s speed in still water =? Let, the swimmer’s speed in still water be x km/h and that of stream be y km/h. According to question, in case of downstream, x + y = 20 ………………… (i) also, in case of upstream, x – y = 8 ………………….. (ii) Now, solving these equations, we get, On adding, 2x = 28 Or, [latex] x = \frac{28}{2} = 14\ km/h [/latex] Putting the value of x in equation (i), we get, 14 + y = 20 Or, y = 6 km/h Hence, swimmer’s speed in still water = 14 km/h And, the velocity of stream = 6 km/h.
A body is projected upwards making an angle [latex]\theta[/latex] with the horizontal with a velocity of 300 ms-1. Find the value of [latex]\theta[/latex] so that the horizontal range will be maximum. Hence find its range and time of flight. Solution Angle of projection [latex](\theta)[/latex] =? for maximum horizontal range. Velocity (u) = 300 m/s Horizontal range (Rmax) =? And, time of flight (T) =? We know that, [latex] \text{Horizontal range (R)} = \frac{u^2sin2\theta}{g} [/latex] This will be maximum if [latex] Sin2\theta = 1 [/latex] [latex] Or, Sin2\theta = Sin90^o [/latex] [latex] Or, 2\theta = 90^o [/latex] [latex] Or, \theta = 45^o [/latex] So, angle of projection is 45o. Now, Maximum horizontal range [latex] (R_{max}) = \frac{u^2}{g} [/latex]
[latex] = \frac{300^2}{10} [/latex]
[latex] = \frac{90000}{10} = 9000 m [/latex] Also, [latex] \text{time of flight (T)} = \frac{2usin\theta}{g} [/latex]
A stone on the edge of a vertical cliff is kicked so that its initial velocity is 9 m/s horizontally. If the cliff is 200 m high, calculate: i. Time taken by stone to reach the ground. ii. How far from the cliff the stone will hit the ground? Solution Initial velocity (u) = 9 m/s (Horizontally) Height of the cliff (h) = 200 m Time taken by stone to reach the ground (t) =? Also, horizontal range of the stone (R) =? We know that, for vertical motion, [latex] h = ut + \frac{1}{2}gt^2 [/latex] [latex] Or, h = \frac{1}{2}gt^2 [/latex] [∵ for vertical motion, u = 0] [latex] Or, 200 = \frac{1}{2}gt^2 [/latex] [latex] Or, 200\times 2 = 10t^2 [/latex] [latex] Or, t^2 = \frac{400}{10} [/latex] Or, t2 = 40 [latex] Or, t = \sqrt{40} = 6.32\ sec[/latex] Now, for horizontal range (R) = u x t = 9 x 6.32 = 56.92 m
