Given data
*The intial horizontal velocity of the potato is u_x = v_o
*The vertical acceleration of the potato is a_y = -g
*The given angle is 45 degree
(a)
The formula to calculate the horizontal co-ordinate of potato is given as
[tex]x=u_xt+\frac{1}{2}a_xt^2[/tex]*Here a_x = 0 m/s^2 is the horizontal acceleration of the potato
Substitute the known values in the above expression as
[tex]\begin{gathered} x=v_0t+\frac{1}{2}(0)(t^2) \\ =v_0t \end{gathered}[/tex]The formula to calculate the vertical co-ordinate of potato is given as
[tex]y=u_yt+\frac{1}{2}a_yt^2[/tex]*Here u_y = 0 m/s is the initial vertical velocity of the potato
Substitute the known values in the above expression as
[tex]\begin{gathered} y=(0)(t)+\frac{1}{2}(-g)t^2 \\ =-\frac{1}{2}gt^2 \end{gathered}[/tex]As from the given data, apply the condition as
[tex]\begin{gathered} x=y \\ v_0t=\frac{1}{2}gt^2 \\ t=\frac{2v_0}{g} \end{gathered}[/tex]Hence, the equation for the time taken is t = 2v_0/g
The horizontal co-ordinate of potato is calculated as
[tex]\begin{gathered} x=v_0t \\ =v_0(\frac{2v_0}{g})_{} \\ =\frac{2v^2_0}{g} \end{gathered}[/tex]The vertical coordinate of the potato is calculated as
[tex]\begin{gathered} y=-\frac{1}{2}gt^2 \\ =-\frac{1}{2}g(\frac{2v_0}{g})^2 \\ =-\frac{2v^2_0}{g} \end{gathered}[/tex]Hence, the co-ordinate of the potato is (2v_0^2/g, -2v_0^2/g)
(b)
The formula to calculate the angle of position co-ordinates is given as
[tex]\tan \theta=\frac{y}{x}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \tan 45^0=\frac{y}{x} \\ x=y \end{gathered}[/tex]The above equation shows that time taken is the same for the angle and co-ordinate is also the same.
Thus, the equation for the time taken is the same as t = 2v_0/g, and the co-ordinate of the potato is the same as (2v_0^2/g, -2v_0^2/g)
