Answer:
39.123 m/s (3 d.p.)
Explanation:
When a body is projected through the air with initial speed u, at an angle of θ to the horizontal, it will move along a curved path. Therefore, trigonometry can be used to resolve the body's initial velocity (u) into its vertical and horizontal components:
In this case, as the baseball is hit an angle of 36.1° to the horizontal then:
To find the initial speed of the ball (u), first find an expression for the time (t) it took for the ball to clear the wall by resolving horizontally.
In this case:
[tex]\begin{aligned}\textsf{Using:}\quad s&=\dfrac{1}{2}(u+v)t\\\\\implies 130.7&=\dfrac{1}{2}(u\cos 36.1^{\circ}+u\cos 36.1^{\circ})t\\\\130.7&=(u\cos 36.1^{\circ})t\\\\t&=\dfrac{130.7}{u\cos 36.1^{\circ}}\end{aligned}[/tex]
Now, find an expression in terms of t by resolving vertically, taking up as positive.
In this case:
[tex]\begin{aligned}\textsf{Using:}\quad s &= ut + \dfrac{1}{2}at^2\\\\7 &= (u \sin 36.1^{\circ})t + \dfrac{1}{2}(-9.81)t^2\\\\7 &= (u \sin 36.1^{\circ})t -4.905t^2\end{aligned}[/tex]
Substitute in the expression we found for t:
[tex]7 = (u \sin 36.1^{\circ})\left(\dfrac{130.7}{u\cos 36.1^{\circ}}\right) -4.905\left(\dfrac{130.7}{u\cos 36.1^{\circ}}\right)^2[/tex]
Simplify:
[tex]7 =130.7\tan 36.1^{\circ}-4.905\left(\dfrac{130.7}{u\cos 36.1^{\circ}}\right)^2[/tex]
[tex]4.905\left(\dfrac{130.7}{u\cos 36.1^{\circ}}\right)^2 =130.7\tan 36.1^{\circ}-7[/tex]
[tex]\left(\dfrac{130.7}{u\cos 36.1^{\circ}}\right)^2 =\dfrac{130.7\tan 36.1^{\circ}-7}{4.905}[/tex]
[tex]\dfrac{1}{u^2}\cdot \left(\dfrac{130.7}{\cos 36.1^{\circ}}\right)^2 =\dfrac{130.7\tan 36.1^{\circ}-7}{4.905}[/tex]
Multiply all terms by u²:
[tex]\left(\dfrac{130.7}{\cos 36.1^{\circ}}\right)^2 =\left(\dfrac{130.7\tan 36.1^{\circ}-7}{4.905}\right)u^2[/tex]
[tex]u^2=\dfrac{\left(\dfrac{130.7}{\cos 36.1^{\circ}}\right)^2}{\left(\dfrac{130.7\tan 36.1^{\circ}-7}{4.905}\right)}[/tex]
[tex]u^2=1453.375722...[/tex]
[tex]u = 38.1231651...[/tex]
Therefore, the initial speed of the ball was 39.123 m/s (3 d.p.).
