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Sagot :
Given,
The distance between the cannon and the wall, x=13.5 m
The height of the wall, h₀=7.9 m
The angle of projection, θ=51.9°
The time duration in which the ball passes the point that is directly above the wall, t=1.15 s
A)
The x-component of the initial velocity is given by,
[tex]\begin{gathered} u_x=\frac{x}{t} \\ \Rightarrow u\times\cos \theta=\frac{x}{t} \\ \Rightarrow u=\frac{x}{t\times\cos \theta} \end{gathered}[/tex]Where u is the initial velocity of the cannon ball.
On substituting the known values,
[tex]\begin{gathered} u=\frac{13.5}{1.15\times\cos51.9^{\circ}} \\ =19.03\text{ m/s} \end{gathered}[/tex]Thus the speed with which the cannon ball was fired is 19.03 m/s
B)
The height of the cannonball at the point that is directly above the wall is given by,
[tex]h=u\sin \theta\times t-\frac{1}{2}gt^2[/tex]Where g is the acceleration due to gravity.
On substituting the known values.
[tex]\begin{gathered} h=19.03\times\sin 51.9^{\circ}\times1.15-\frac{1}{2}\times9.8\times1.15^2 \\ =10.74\text{ m} \end{gathered}[/tex]Thus, the vertical distance by which the cannon ball clears the top of the wall is,
[tex]\begin{gathered} H=h-h_0 \\ =10.74-7.9 \\ =2.84\text{ m} \end{gathered}[/tex]Thus the cannon ball clears the top of the wall by a distance of 2.84 m
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