Customised for the Numbas demo exam
\nMotion under gravity. Object is projected vertically with initial velocity $V\\;m/s$. Find time to maximum height and the maximum height. Now includes an interactive plot.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "A Numbas question can include interactive graphics, such as this plot of the trajectory given by the student's answer.
\nSee this question in the public editor
\nA ball is thrown upwards, and moves according to the equation $\\displaystyle{\\frac{d^2z}{dt^2}=-g}$
(where $z(t)$ is distance in metres measured upwards from the ground and the constant acceleration of gravity, $g$ , is given as $9.81\\;m/s^2$).
The ball is projected upwards with a speed $\\var{v}\\;m/s$.
", "advice": "Integrating $\\displaystyle{\\frac{\\mathrm{d}^2z}{\\mathrm{d}t^2}=-g}$ once gives the velocity $\\displaystyle{\\frac{\\mathrm{d}z}{\\mathrm{d}t}=-gt+A}$.
\nBut $A=\\var{v}$ as the velocity is $\\var{V}\\;m/s$ at $t=0$.
\nSo the velocity is
\n\\begin{align} \\frac{\\mathrm{d}z}{\\mathrm{d}t} &= \\var{v}-gt & (1) \\end{align}
\nIntegrating again gives
\n\\[ z = \\var{v}t-\\frac{g}{2}t^2+B \\]
\nand $B=0$ as $z=0$ at $t=0$.
\nHence the distance travelled upwards is given by
\n\\begin{align} z &= \\var{v}t-\\frac{g}{2}t^2 & (2) \\end{align}
\n{advicegraph()}
\nThe time $t_{\\text{max}}$ taken to reach maximum height is the time satisfying $\\displaystyle{\\frac{dz}{dt}=0}$
\n$t_{\\text{max}}$ is given from equation $(1)$ by
\n\\begin{align}
\\var{v} - gt_{\\text{max}} &= 0 \\\\
gt_{\\text{max}} &= \\var{v} \\\\
t_{\\text{max}} &= \\frac{\\var{v}}{g} \\\\[0.5em]
&= \\frac{\\var{v}}{9.81} \\\\[0.5em]
&= \\var{t1}
\\end{align}
(to $2$ decimal places)
\nThis is at the point $A$ in the graph above.
\nThe maximum height $z_{\\text{max}}$ is given from equation $(2)$ by substituting in the value $t_{\\text{max}}= \\var{v}/g$, giving
\n\\begin{align}
z_{\\text{max}} &= \\var{v} \\times \\frac{\\var{v}}{g} - \\frac{g}{2}\\left(\\frac{\\var{v}}{g}\\right)^2 \\\\
&= \\frac{\\var{v}^2}{g}-\\frac{g\\var{v}^2}{2g^2} \\\\
&= \\frac{\\var{v}^2}{2g} \\\\
&= \\var{mh}\\;m
\\end{align}
(to $2$ decimal places)
\nThis is at the point $B$ in the graph above.
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\nInput the vertical distance $z$ as a function of $t$.
\nNote that at $t=0$ we have $z=0$ and that $\\displaystyle \\frac{dz}{dt}=\\var{v}m/s$.
\nInput gravitational acceleration as $g$.
\n$z=$ [[0]]
\nYour formula is plotted in the graph above. The vertical axis represents $z$ and the horizontal axis represents $t$.
\nNote that the blue line indicates that:
\nTime taken to reach maximum height = [[0]] $s$ (accurate to $2$ decimal places)
\nMaximum height = [[1]] $m$ (accurate to $2$ decimal places)
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