First slide

Rectilinear Motion

Question

The maximum height attained by projectile is increased by 10% by changing the angle of projection, without changing the speed of projection. The percentage increase in the time of flight will be

Moderate

A

20%
B

15%
C

10%
D

5%

Solution

Given $h = \frac{u^{2} \sin^{2} θ}{2 g}$ .
Differentiating partially we get $(∵ v_{0} = constant)$
$δh = \frac{u^{2}}{2 g} 2 \sin θ \cos θ δθ$ .
Thus $\frac{δh}{h} = \frac{2 \cos θδθ}{\sin θ} = 0 .1$ (given).
Therefore, $\frac{\cos θδθ}{\sin θ} = 0 .05$ ……(1)
We also have $T = \frac{2 usin θ}{g}$ which gives $δT = \frac{2 ucos θδθ}{g}$ .
Thus $\frac{δT}{T} = \frac{\cos θδθ}{\sin θ}$ .
But using (1) we get
$\frac{δT}{T} = 0 .05 or δT = 0 .05 T$ .
Hence, T increases by 5%.

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