Projectile Motion Question (1 Viewer)

frmldhyd · Jul 1, 2011

A bullet is fired horizontally with a velocity of 800 from a height of 3m above the the horizontal plane. What is the range of the bullet?

Drongoski · Jul 1, 2011

velocity 800 ms^-1?

x(t) = 800t

$\ddot y(t) = -g$

$\therefore \dot y(t) = \int -g dt = -gt + C$

$\dot y(0) = 0 \implies C = 0$

$\therefore \dot y(t) = -gt$

$\therefore y(t) = \int -gt dt = -\frac{1}{2}gt^2 + D$

$y(0) = 3 \implies D = 3 \implies y(t) = -\frac {1}{2}gt^2 + 3$

$y(t) = 0 \implies t = \sqrt \frac {6}{g}$

$range = 800 \times \sqrt \frac {6}{9.8} = 625.9686 . . m$

hscishard · Jul 1, 2011

625.97m @-9.8
619.68m @-10

Projectile Motion Question (1 Viewer)

frmldhyd

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Drongoski

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hscishard

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