Mastering Physics: Arrow Hits Apple
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Question “Mastering Physics: Arrow Hits Apple”
An arrow is shot at an angle of above thehorizontal. The arrow hits a tree a horizontal distance away, at the sameheight above the ground as it was shot. Use for themagnitude of theacceleration due to gravity.
Suppose someone drops an apple from a vertical distance of, directly abovethe point where the arrow hits the tree.
How long after the arrow was shot should theapple be dropped, in order for the arrow to pierce the apple as thearrow hits the tree?Express your answernumerically in seconds, to two significant figures.
t_d=?s
I got the previous answer right but this one stumps me for reason.It seems simple but I can’t grasp it entirely. Would someone pleasework it out step by step so I cansee what I should do in thesekinds of situations?
Answer
This problem can be solved by focusing on horizontal range and flight time for projectiles that are projected at an angle above the ground.
First, calculate how long the arrow stays in the air. Next, calculate the time it takes for the arrow to descend. Calculate the time it takes for an apple to fall from a tree.
This formula can calculate the horizontal range for a projectile that is projected at an angle above ground.
D = \fracV_0^2\sin \left( 2\theta \right)gHere, V_0
Rearrange the equation forV_0 V_0 = \sqrt \fracDg\sin \left( 2\theta \right)
This formula can calculate the time it takes for a projectile to fly if it is projected at an angle above the ground.
T = \frac2V_0\sin \theta g(a)
Substitute 220m for D, 9.8\;{\rm{m/}}{{\rm{s}}^2} \beginarrayc\\V_0 = \sqrt \frac220\;\rmm\left( 9.8\;\rmm/\rms^2 \right)\sin \left( 2\left( 45^\circ \right) \right) \\\\ = 46.43\;\rmm/s\\\endarray
Substitute 46.43m/s V, 9.8\;{\rm{m/}}{{\rm{s}}^2} \beginarrayc\\t_a = \frac2\left( 46.43\;\rmm/s \right)\sin \left( 45^\circ \right)9.8\;\rmm/\rms^2\\\\ = 6.7\;\rms\\\endarray
(b)
The useful kinematic equation for motion is the relationship between distance traveled ( s), initial speed ( u), acceleration and t time.
s = V_0t + \frac12at^2Substitute 0 forV_0 \begin{array}{r}\\h = 0 + \frac{1}{2}g{t^2}\\\\{t^2} = \frac{{2h}}{g}\\\\t = \sqrt {\frac{{2h}}{g}} \\\end{array}
Substitute 6m for h, and 9.8\;{\rm{m/}}{{\rm{s}}^2} \begin{array}{c}\\{t_d} = \sqrt {\frac{{2\left( {6\;{\rm{m}}} \right)}}{{\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}} \\\\ = 1.11\;{\rm{s}}\\\end{array}
It takes,
\begin{array}{c}\\t = {t_a} - {t_d}\\\\ = 6.7\;{\rm{s - 1}}{\rm{.11s}}\\\\{\rm{ = 5}}{\rm{.59}}\;{\rm{s}}\\\end{array}It takes 5.59 seconds to remove the apple from its tree.
Ans: Part A
The time the arrow spends in air is 6.7 seconds.
Part b
It takes 5.59 seconds to remove the apple from its tree.
Conclusion
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