A ball is projected upwards from top of tower with a velocity 50,,m{s^{ - 1}} making an angle {30^o}

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3-2.Motion in Plane

hard

A ball is projected upwards from the top of tower with a velocity $50\,\,m{s^{ - 1}}$ making an angle ${30^o}$ with the horizontal. The height of tower is $ 70 \,m$. After how many seconds from the instant of throwing will the ball reach the ground ........ $\sec$

A

$2$

B

$5$

C

$7$

D

$9$

Solution

(c) The vertical component of velocity of projection $ = – 50\sin 30^\circ = – 25\,m/s$

If $t$ be the time taken to reach the ground,

$h = ut + \frac{1}{2}g{t^2}$

$⇒$ $70 = – 25t + \frac{1}{2} \times 10{t^2}$

$⇒$ $70 = – 25t + 5{t^2}$

$⇒$ ${t^2} – 5t – 14 = 0$

$⇒$ $t= -2s$ and $7s$

Since, $t = -2s$ is not valid

$\therefore t = 7\, s$

Standard 11

Physics

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