Two and Three dimensional motion

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a bullet shot at an initial speed \( v_0 \) and angle \( \theta \). Key equations used include the horizontal motion equation \( x = x_0 + v_0 \cos(\theta) t \) and the vertical motion equation \( y = y_0 + v_0 \sin(\theta) t - 5t^2 \), with gravitational acceleration \( g \) set to -10 m/s². The maximum height is determined by setting the derivative of the vertical position to zero, while the total time of flight is calculated as \( t = \frac{v_0 \sin(\theta)}{10} \). The horizontal distance traveled is derived from the total time of flight.

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Homework Statement



A bullet is shot with a speed v0 at an angle θ above the horizontal.
a) what is its maximum height?
b) how long does it take before it hits the earth?
c) where does it strike the earth?

Homework Equations





The Attempt at a Solution


x component: x= x0 + uxt + 1/2axt^2
= v0cosθt

y component: y= y0 + uyt + 1/2ayt^2
= v0sinθt - 5t^2
assume that g=-10

a) when maximizing y, dy/dt=0=v0sinθt - 10t

b) set y = 0 --> 0= v0sinθt - 10t
0 = t(v0sinθ - 10t)
t = 0, t= v0sinθ/10

c) maximize x, I assumed that dx/dt = 0. But I'm not too sure on this part.
 
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you have the total time of flight as t=(v0sinθ)/10. So what the total 'x' distance it travels?
 

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