- 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]
[latex]
= \frac{2\times 300 \times sin45^o}{10}
[/latex]
[latex]
= \frac{600\times \frac{1}{\sqrt{2}}}{10}
[/latex]
[latex]
= \frac{60}{\sqrt{2}} = 42.43 s
[/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
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