[填空题]
A projectile is launched a)u ojt8y+api tbjy; )1zuxu -0t an kr8) 8 309fhw mxonqw4m:l faoangle of $30^{\circ}$ to the horizontal with a speed of $ 30 \mathrm{~m} / \mathrm{s}$. How does the horizontal component of its velocity 1.0 s after launch compare with its horizontal component of velocity 2.0 s after launch?
参考答案: 26±2%
本题详细解析: The horizontal component of the velocity stays constant in projectile xr2q)vqux 0x)motion, assuming that air resistance is negligible. Thus the horizontal component of velocity 1.0 seconds after launch will be the same as the horizontal component of 2 velocity 2.0 seconds after launch. In both cases the horizontal velocityxx2q)urv q) x0 will be given by $ v_{x}=v_{0} \cos \theta=(30 \mathrm{~m} / \mathrm{s})\left(\cos 30^{\circ}\right)=26 \mathrm{~m} / \mathrm{s} $.
