Question

1.) A projectile is launched at 15 degrees. The landing height is the same as the launch position. At what other angle can the projectile be launch to achieve the same horizontal displacement?
2.) A football is kicked into the air with a velocity of 32m/s at an angle of 25º. At the very top of the ball’s path, its vertical velocity is?

Answer 1

1) The general equations of motion of the projectile on the x and y axis are:
$x(t) = v_0 \cos \alpha t$
$y(t)=v_0 \sin \alpha t - \frac{1}{2}gt^2$
where v0 is the initial velocity, $\alpha$ is the angle with respect to the ground, and $g=9.81 m/s^2$ is the gravitational acceleration. We can see that the motion of the projectile is an uniform motion on the x-axis and an uniformly accelerated motion on the y-axis.

First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of $15^{\circ}$. To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
$v_0 \sin \alpha t - \frac{1}{2}gt^2 =0$
$t( v_0 \sin \alpha - \frac{1}{2} gt)=0$
that has two solutions: t=0 (beginning of the motion) and
$t= \frac{2 v_0 \sin \alpha}{g}$
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
$x_1=v_0 \cos \alpha t = v_0 \cos \alpha ( \frac{2 v_0 \sin \alpha }{g} )= \frac{2 v_0^2}{g} \sin \alpha \cos \alpha$
with $\alpha=15^{\circ}$.

If we call $\beta$ the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
$x_2 = \frac{2 v_0^2}{g} \sin \beta \cos \beta$
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
$\sin \alpha \cos \alpha = \sin \beta \cos \beta$
Now let's remind that $\cos \theta= \sin (90^{\circ} -\theta)$ so that we can rewrite the equation as
$\sin \alpha \sin (90^{\circ}-\alpha) = \sin \beta \sin (90^{\circ}-\beta)$
and using $\alpha=15^{\circ}$:
$\sin 15^{\circ} \sin (75^{\circ}) = \sin \beta \sin (90^{\circ}-\beta)$
and we can see that there are two values of $\beta$ that satisfy the equation: $\beta=\alpha=15^{\circ}$ and $\beta=75^{\circ}$, which is the solution of our problem.

2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.