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Let us draw a sketch to understand the situation
We will use some rules here
[tex]\begin{gathered} v_x=vcos\theta=64cos(34) \\ d_x=v_xt=64cos(34)t \end{gathered}[/tex]Since the horizontal distance is 50 yards
Since 1 yard = 3 feet, then
[tex]d_x=50\times3=150feet[/tex]We will use it to find the time t
[tex]\begin{gathered} d_x=150 \\ 64cos(34)t=150 \\ t=\frac{150}{64cos(34)}\text{ s} \end{gathered}[/tex]Now, we will find the vertical distance (h) by using this rule
[tex]\begin{gathered} v_y=vsin\theta=64sin(34) \\ d_y=h=v_yt-\frac{1}{2}at^2=64sin(34)t-\frac{1}{2}(32)t^2 \end{gathered}[/tex]Note that: a is the acceleration of gravity which is 32 ft/s^2
We will substitute t by its value
[tex]h=64sin(34)(\frac{150}{64cos(34)})-16(\frac{150}{64cos(34)})[/tex]We can simplify it by using sin34/cos34 = tan34, and 1/cos34 = sec34
But I will put it on the calculator to find the final answer
[tex]h=55.94\text{ ft}[/tex]Since the height of the crossbar is 10 feet, then
Sanjay's attempt successful
