Question

With what minimum speed must a particle be projected from origin so that it is able to pass through a given point $(30m,40m)$. Take $g=10m/s^2$.

• A. $60m/s$
• B. $50m/s$
• C. $30m/s$
• D. $40m/s$

Answer 1

The correct option is C $30m/s$

From the above figure,

Along x-axis,

$30=u_{x}t$

$30=ucos\theta \times t$

$t=\frac{30}{ucos\theta }$. . . . . . . . . (1)

Along y-axis,

$40=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$40=usin\theta \times t-\frac{1}{2}g{t}^2$. . . . . . . . . . . .(2)

Substitute equation (1) in equation (2), we get

$40=\frac{usin\theta \times 30}{ucos\theta }-\frac{1}{2}\times 10\times (\frac{30}{ucos\theta }{)}^2$

$40=30tan\theta -\frac{4500}{u^{2}co{s}^2\theta }$

$40-30tan\theta =-\frac{4500}{u^2}se{c}^2\theta$

$(40-30tan\theta )=[-\frac{4500}{u^2}(1+ta{n}^2\theta \left)\right]$

By squaring both side,

$(40-30tan\theta {)}^2=[-\frac{4500}{u^2}(1+ta{n}^2\theta ){]}^2$

$1600+900ta{n}^2\theta -2400tan\theta =\frac{(4500{)}^2}{u^4}(1+ta{n}^2\theta {)}^2$

$u^4=\frac{\left(4500{)}^2\right(1+ta{n}^2\theta {)}^2}{(40-30\tan \theta {)}^2}$

$u^2=4500\left(\frac{1+ta{n}^2\theta }{40-30tan\theta }\right)$

For minimum speed,

$u^2=\frac{4500}{5}=900$

$u=\sqrt{900}m/s$

$u=30m/s$

The correct option is C.