Consider the motion of the following objects. Assume the x-axis is horizontal, the positive y-axis is vertical, the ground is horizontal, and only the gravitational force acts on the object. A projectile is launched from a platform 20ft20 \mathrm{ft} above the ground at an angle of 6060^{\circ} above the horizontal with a speed of 250ft/s250 \mathrm{ft} / \mathrm{s}. Assume the origin is at the base of the platform. Determine the time of flight and range of the object.


Answered 6 months ago
Answered 6 months ago
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Projectile lands on the ground for some value of t>0t>0, therefore to determine the time of flight we need to solve y(t)=0y(t)=0.

y(t)=016t2+1253t+20=0t1,2=1253±(1253)24(16)202(16)t1,2=1253±4815532t1,2216.25±219.4432t435.693213.6 s,  t>0\begin{align*} y(t)&=0 \\ -16t^2+ 125\sqrt{3}t +20 &=0 \\ t_{1,2}&=\dfrac{-125\sqrt{3} \pm \sqrt{(125\sqrt{3})^2 - 4\cdot (-16) \cdot 20 }}{2\cdot (-16)}\\ t_{1,2}&=\dfrac{-125\sqrt{3} \pm \sqrt{48155}}{-32}\\ t_{1,2}& \approx \dfrac{-216.25 \pm 219.44}{-32}\\ t&\approx \frac{435.69}{32} \approx 13.6 \ \mathrm{s}, \ \ t>0 \end{align*}

To determine the range of the projectile we must include the obtain time into horizontal distance x(t)=125tx(t)=125t. Therefore, x(13.6)1702.9 ftx(13.6) \approx 1702.9 \ \mathrm{ft}.

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