Answer to Question #247030 in Quantum Mechanics for Elli

Question

 What is the range of the shuttlecock launched with an initial speed of 6.3797 m/s at a 58.3699-degree angle with respect to the horizontal from a height of 1.8914 m in a planet where the downward gravitational acceleration has a magnitude of 11.8201 m/s²?

Answer 1

$l=l_1+l_2,$

$l_1=\frac{v^2\sin2\alpha}g=3.0751~m,$

$h=v\sin\alpha t+\frac{gt^2}2,\implies$

$t=\frac{\sqrt{v^2\sin^2\alpha+2gh}-v\sin\alpha}g=0.2693~s,$

$l_2=v\cos\alpha t=0.9010~m,$

$l=3.9761~m.$