A bomb is fired horizontally with a velocity of $20 m s^{- 1}$ from the top of a tower $40$ m high. After how much time and at what horizontal distance from the tower will the bomb strike the ground? Take $g = 9.8 m s^{- 2}$ .

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Solution

$h = \frac{1}{2} 10 (3)^{2} = 45 m$
$v cos 30^{o} = 10 \times 3$
$\Rightarrow v = 20 \sqrt{3} m s^{- 1}$
$u = v sin 30^{o}$
$= 20 \sqrt{3} \times \frac{1}{2} = 10 \sqrt{3} m s^{- 1}$

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A bomb is fired horizontally with a velocity of 20ms−1 from the top of a tower 40 m high. After how much time and at what horizontal distance from the tower will the bomb strike the ground? Take g=9.8ms−2.

h=1210(3)2=45mvcos30o=10×3⇒v=20√3ms−1u=vsin30o=20√3×12=10√3ms−1

A bomb is fired horizontally with a velocity of $20 m s^{- 1}$ from the top of a tower $40$ m high. After how much time and at what horizontal distance from the tower will the bomb strike the ground? Take $g = 9.8 m s^{- 2}$ .

$h = \frac{1}{2} 10 (3)^{2} = 45 m$
$v cos 30^{o} = 10 \times 3$
$\Rightarrow v = 20 \sqrt{3} m s^{- 1}$
$u = v sin 30^{o}$
$= 20 \sqrt{3} \times \frac{1}{2} = 10 \sqrt{3} m s^{- 1}$