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daser333 [38]
3 years ago
6

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

s the projectile hit the ground and with what speed?
Mathematics
1 answer:
qaws [65]3 years ago
7 0

The projectile's horizontal and vertical positions at time t are given by

x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t

y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2

where g=9.8\dfrac{\rm m}{\mathrm s^2}. Solve y=0 for the time t it takes for the projectile to reach the ground:

30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s

In this time, the projectile will have traveled horizontally a distance of

x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m

The projectile's horizontal and vertical velocities are given by

v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ

v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about \sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}.

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