Question

A particle is thrown with velocity $u$ making an angle $\theta$ with the vertical. It just crosses the top of two poles each of height $h$ after $1s$ and $3s$ respectively. The maximum height of projectile is-

• A. $9.8m$
• B. $19.6m$
• C. $39.2m$
• D. $4.9m$

Answer 1

The correct option is B $19.6m$

times taken to reach P is $1s$ therefore

$v^2=u^2+2aS$

$v^2=u^{2}co{s}^2\theta -2gh$

let the distance from P to reach at the top be x

$v^2=u^2+2as$

$0=u^{2}co{s}^2\theta -2gh-2gx$

$x=\frac{u^{2}co{s}^2\theta }{2g}-h$

time taken is $v=u+at$

$0=ucos\theta -g-gt$

$t=\frac{ucos\theta }{g}-1$

time taken to reach top from P is $1s$

$2g=ucos\theta$

maximum height $H=\frac{u^{2}co{s}^2\theta }{2g}$ as the verical component is $ucos\theta$

therefore $H=\frac{(2g{)}^2}{2g}=2g=19.6m$