Answer to Question #236150 in Calculus for mika

Question

An object thrown from a height of 2 m above the ground follows a parabolic path

until the object falls to the ground; see Figure. If the object reaches a maximum

height (measured from the ground) of 7 m after travelling a horizontal distance of 4

m, determine the horizontal distance between the object's initial and final positions.

Answer 1

The motion can be broken into horizontal and vertical motions

$$vertical: y(t)=h_0+v_{0y}t-\dfrac{gt^2}{2}$$

$$horizontal : x(t)=v_{0x}t$$

The highest point in any trajectory, called the apex, is reached when

$$v_y(t_1)=v_{0y}-gt_1=0$$

Then

$$y(t_1)=h_{max}=h_0+v_{0y}(\dfrac{v_{0y}}{g})-\dfrac{g}{2}(\dfrac{v_{0y}}{g})^2=h_0+\dfrac{v_{0y}^2}{2g}$$

$$=h_0+\dfrac{v_{0y}^2}{2g}$$

$$v_{0y}=\sqrt{2g(h_{max}-h_0)}$$

$$v_{0y}=\sqrt{2(9.8\ m/s^2)(7\ m-2\ m)}=7\sqrt{2}\ m/s$$

$$x(t_1)=v_{0x}t_1=\dfrac{v_{0x}v_{0y}}{g}$$

$$v_{0x}=\dfrac{4\ m(9.8\ m/s^2)}{7\sqrt{2}\ m/s}=2.8\sqrt{2}\ m/s$$

The object will fall to the ground in time $t_2$

$$y(t_2)=h_0+v_{0y}t_2-\dfrac{gt_2^2}{2}=0$$

$$2+7\sqrt{2}t_2-\dfrac{9.8t_2^2}{2}=0, t_2>0$$

$$4.9t_2^2-7\sqrt{2}t_2-2=0$$

$$D=(-7\sqrt{2})^2-4(4.9)(-2)=137.2$$

$$t_2=\dfrac{7\sqrt{2}\pm\sqrt{137.2}}{2(4.9)}=\dfrac{\sqrt{2}\pm\sqrt{2.8}}{1.4}$$

Since $t_2>0,$ then we take

$$t_2=\dfrac{\sqrt{2}+\sqrt{2.8}}{1.4}$$

The horizontal distance between the object's initial and final positions is

$$d=x(t_2)-x(0)=v_{0x}t_2-0$$

$$d=2.8\sqrt{2}(\dfrac{\sqrt{2}+\sqrt{2.8}}{1.4})=4(1+\sqrt{1.4}) (m)$$