Đáp án:

Giải thích các bước giải:

 $x=v_0\text{cos}\ \alpha\ t$

$y=v_0\text{sin}\ \alpha\ t-\dfrac{1}{2}gt^2$

Chạm mặt bàn khi $y=h⇒\dfrac{g}{2}t^2-v_0\text{sin}\ \alpha\ t+h=0$

Gọi $2$ nghiệm $t_1<t_2$

$⇒Δ=v^2_0\text{sin}^2\ \alpha-2gh$

$t_2-t_1=\dfrac{\sqrt{Δ}}{\dfrac{g}{2}}=\dfrac{2}{g}\sqrt{v^2_0\text{sin}^2\ \alpha-2gh}$

$⇒AB=x(t_2)-x(t_1)=v_0\text{cos}\ \alpha (t_2-t_1)$

$⇒AB=\dfrac{2v_0\text{cos}\ \alpha}{g}\sqrt{v^2_0\ \text{sin}^2\ \alpha-2gh}$

$AB$ lớn nhất$⇔AB^2$ lớn nhất

$⇒AB^2=\dfrac{4v^2_0}{g^2}\ \text{cos}^2\ \alpha\left(v^2_0\ \text{sin}^2\ \alpha-2gh\right)$

$⇒AB^2=\dfrac{160}{100}\ \left(1-\text{sin}^2\ \alpha\right)\left(40\ \text{sin}^2\ \alpha-20\right)$

Đặt $u=\text{sin}^2\ \alpha\left(≥\dfrac{1}{2}\right)$

$AB^2=1,6(-40u^2+60u-20)$

$⇒AB^2=1,6\left(\dfrac{5}{2}-40\left(u-\dfrac{3}{4}\right)^2\right)$

$→AB$ lớn nhất khi $u=\dfrac{3}{4}$

$⇒\text{sin}^2\ \alpha=\dfrac{3}{4}⇒\text{sin}\ \alpha=\dfrac{\sqrt{3}}{2}$

$⇒\alpha=60^\circ$