## Related questions with answers

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 $20 \mathrm{ft}$ above the ground at an angle of $60^{\circ}$ above the horizontal with a speed of $250 \mathrm{ft} / \mathrm{s}$. Assume the origin is at the base of the platform. Determine the time of flight and range of the object.

Solution

VerifiedProjectile lands on the ground for some value of $t>0$, therefore to determine the time of flight we need to solve $y(t)=0$.

$\begin{align*} y(t)&=0 \\ -16t^2+ 125\sqrt{3}t +20 &=0 \\ t_{1,2}&=\dfrac{-125\sqrt{3} \pm \sqrt{(125\sqrt{3})^2 - 4\cdot (-16) \cdot 20 }}{2\cdot (-16)}\\ t_{1,2}&=\dfrac{-125\sqrt{3} \pm \sqrt{48155}}{-32}\\ t_{1,2}& \approx \dfrac{-216.25 \pm 219.44}{-32}\\ t&\approx \frac{435.69}{32} \approx 13.6 \ \mathrm{s}, \ \ t>0 \end{align*}$

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

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