91Ó°ÊÓ

Given initial speed $v_0=40m/s$and height of 300 ft. with an angle$\mathrm{Î\pm }={45}^{∘}$

We have $x=\left(v_0\cos \mathrm{θ}\right)t$

where $x$represent how far the object travel in time $t.$

role="math" localid="1649009028138" $x=\left(40\cos {45}^{∘}\right)t$

and $y=-\frac{1}{2}g{t}^2+\left(v_0\sin \mathrm{θ}\right)t+h$

where $y$represent the height of the object

$$\begin{gathered} y=-\frac{1}{2}\left(9.8\right)t^2+\left(40\sin {45}^{∘}\right)t+300 \\ y=-4.9t^2+\left(40\sin {45}^{∘}\right)t+300 \end{gathered}$$

Let us set the height $y=0$

$y=-4.9t^2+\left(40\sin {45}^{∘}\right)t+300$

$$\begin{gathered} t=\frac{\left(-40\sin 45\right)Â\pm \sqrt{{\left(\left(-40\sin 45\right)\right)}^2-4\left(-4.9\right)300}}{2\left(-4.9\right)} \\ t=\frac{-28.28Â\pm 81.73}{-9.8} \\ t=\frac{-28.28+81.73}{-9.8}or\frac{-28.28-81.73}{-9.8} \\ t=-5.45or11.23 \end{gathered}$$

negative is not valid.

Thus the ball was in the air for$11.23sec.$

From $x=\left(40\cos {45}^{∘}\right)t$

$$\begin{gathered} x=\left(40\cos {45}^{∘}\right)11.23 \\ x=317.52 \end{gathered}$$

the horizontal displacement made by ball is$317.52m$

Use $t=-\frac{b}{2a}$

$t=\frac{-\left(40\sin {45}^{∘}\right)}{2\left(-4.9\right)}=2.89$

At $2.89sec.$the ball was at maximum point

From $y=-4.9t^2+\left(40\sin {45}^{∘}\right)t+300$

$$\begin{gathered} y=-4.9{\left(2.89\right)}^2+\left(40\sin {45}^{∘}\right)2.89+300 \\ y=340.82m \end{gathered}$$

Therefore, the maximum height is

Enter the parametric equations

$X_{1T}=40\cos {45}^{∘}t,Y_{1T}=-4.9t^2+\left(40\sin {45}^{∘}\right)t+300$

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$$\begin{gathered} T_{min}=-5.45,T_{,max}=11.23,T_{step}=0.1 \\ {X}_{min}=0,X_{max}=350,X_{step}=1 \\ {Y}_{min}=-120,Y_{,max}=500,Y_{step}=1 \end{gathered}$$

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