An object is going to be launched from the ground with an initial velocity of $v$. It starts a distance $d$ away from a wall of height $h$. Assume that the wall is close enough that any angle $\Theta < 45 \degree$ would not make it over the wall. Find $\Theta$ to maximize the horizontal distance the object travels, and it must pass over the wall? ![[20241014_131952.jpg]] In projectile motion, the horizontal and vertical motion are independent. The horizontal motion is constant velocity and the vertical motion is constant acceleration (acceleration due to gravity). Horizontal Motion $v_x=\frac{d}{t} \tag{1}$ $t=\frac{d}{v_x} \tag{2}$ $t=\frac{d}{v \cdot cos \ \theta} \tag{3}$ Vertical Motion $h=v_{0,y}t \ + \ \frac{1}{2} gt^2 \tag{4}$ Substitute horizontal time into equation 4 $h=v \cdot sin \ \theta \cdot \frac{d}{v \cdot cos \ \theta} \ + \ \frac{1}{2} g \Biggr(\frac{d}{v \cdot cos \ \theta} \Biggr) ^2 \tag{5}$ $h=d \cdot tan \ \theta \ - \ \frac{g d^2}{2v^2 cos^2 \ \theta} \tag{6}$