Question

A projectile is fired from a point on the ground 10m away from a vertical wall and it just clears the wall and another parallel wall. The distance between the walls is 20m, and the height of each wall is 2m. The plane in which the projectile travel is perpendicular to the plane of the wall. Find the angle above the horizontal at which the projectile was fired, and the greatest height above the walls attained by the projectile.

Answer 1

Let the initial velocity of the projectile be $v$ and the launch angle be $\theta$. Let the velocity of the projectile at the wall be $v_w$ and it makes an angle $\alpha$ to the horizontal. Let also the distance between the two walls be $x_w$.

a) Let's write the equations of motion of the projectile in horizontal and vertical directions when the projectile just clears the wall and flies between them:

$$x_w=v_{wx}t=v_wtcos\alpha, (1)$$ $$y=v_wtsin\alpha-\dfrac{1}{2}gt^2. (2)$$

Let's find the $y$-component of the projectile's velocity above the wall (at the maximum height $v_{wy}=0$ ):

$$v_{wy}^2=v_{w0y}^2-2gy,$$ $$0=v_{w0y}^2-2gy,$$ $$v_{w0y}=\sqrt{2gy}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot2\ m}=6.26\ \dfrac{m}{s}.$$

Let's substitute $v_{w0y}$ into the equation for $y$ and find the time that the projectile needs to fly between two walls:

$$4.9t^2-6.26t=0,$$ $$t=\dfrac{6.26}{4.9}=1.27\ s.$$

Then, we can substitute $t$ into the equation for $x_w$ and find the $x$-component of the projectile's velocity above the wall:

$$v_{wx}=\dfrac{x_w}{t}=\dfrac{20\ m}{1.27\ s}=15.75\ \dfrac{m}{s}.$$

Finally, we can find the velocity of the projectile above the wall from the Pythagorean theorem:

$$v_w=\sqrt{v_{wx}^2+v_{wy}^2}=\sqrt{(15.75\ \dfrac{m}{s})^2+(6.26\ \dfrac{m}{s})^2}=16.95\ \dfrac{m}{s}.$$

Then, we can find the angle $\alpha$ from the geometry:

$$\alpha=sin^{-1}(\dfrac{v_{wy}}{v_w})=sin^{-1}(\dfrac{6.26\ \dfrac{m}{s}}{16.95\ \dfrac{m}{s}})=21.67^{\circ}.$$

Then, we can use the kinematic equation to find the initial velocity of the projectile and the launch angle:

$$v_{yw}^2=v_y^2-2gy,$$ $$v_y=\sqrt{v_{wy}^2+2gy},$$ $$v_y=\sqrt{(6.26\ \dfrac{m}{s})^2+2\cdot9.8\ \dfrac{m}{s^2}\cdot2\ m}=8.85\ \dfrac{m}{s}.$$

Thus, $v_y=vsin\theta=8.85\ \dfrac{m}{s}.$ Let's write the equation of projectile's motion in horizontal direction from the moment when it launches to the moment when it reaches the wall:

$$v_x = vcos\theta=v_wcos\alpha=15.95\ \dfrac{m}{s}.$$

Then, we can find the initial velocity from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(15.75\ \dfrac{m}{s})^2+(8.85\ \dfrac{m}{s})^2}=18.1\ \dfrac{m}{s}.$$

Finally, we can find the launch angle:

$$\theta=cos^{-1}(\dfrac{v_x}{v})=cos^{-1}(\dfrac{15.75\ \dfrac{m}{s})^2}{18.1\ \dfrac{m}{s})^2})=29.5^{\circ}.$$

b) We can find the greatest height above the walls attained by the projectile by substituting rise time that projectile takes to reach the maximum height into the equation (2):

$$t_{rise}=\dfrac{t}{2}=\dfrac{1.27\ s}{2}=0.63\ s.$$ $$y_{max}=y_0 + v_wt_{rise}sin\alpha-\dfrac{1}{2}gt_{rise}^2,$$ $$y_{max}=2\ m + 16.95\ \dfrac{m}{s}\cdot0.63\ s\cdot sin21.67^{\circ}-\dfrac{1}{2}\cdot9.8\ \dfrac{m}{s^2}\cdot(0.63\ s)^2=4.0\ m.$$