At what point of a projectile motion are acceleration and velocity perpendicular to each other?

$A$ projectile fired with initial velocity $u$ at an angle $\theta$ has a range $R$. If the initial velocity is doubled at the same angle of projection,then the new range will be:

At what angle of elevation should a projectile be projected with a velocity of $20 \, m/s$ to reach a maximum height of $10 \, m$ (in $^{\circ}$)?

$A$ ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone,which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^{\circ} - \theta_1)$ from the $+x$-direction with a suitable initial speed $v$. For a fixed $v_0$,when $(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ})$,the stone hits the ball after time $T_1$,and when $(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ})$,it hits the ball after time $T_2$. In such a case,$(T_1 / T_2)^2$ is. . . . .

$A$ projectile is fired from horizontal ground with speed $v$ and projection angle $\theta$. When the acceleration due to gravity is $g$,the range of the projectile is $d$. If at the highest point in its trajectory,the projectile enters a different region where the effective acceleration due to gravity is $g^{\prime}=\frac{g}{0.81}$,then the new range is $d^{\prime}=n d$. The value of $n$ is. . . . .

If the maximum height and range of a projectile are $3 \,m$ and $4 \,m$ respectively,then the velocity of the projectile is (Take $g=10 \,ms^{-2}$)