[填空题]
A projectile is launched adn70anq gub4 ba/ 2qu5b /j,ej.fwfq/t an angle of) ydr+hl17gzm0 qifc0 $30^{\circ}$ 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 motb+rc (e9b-nv1( m 0f(kxot 7f rnmz4pywv/q-vion, assuming that air resistance is negligible. Thus the horizontal component of velocity 1.0 seconds after launch will be the same as the -m 4/(zb xfto wnebr7vr-v +yfm 0qv(pn 1c9k(horizontal component of 2 velocity 2.0 seconds after launch. In both cases the horizontal velocity 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} $.
