Explore Westonci.ca, the top Q&A platform where your questions are answered by professionals and enthusiasts alike. Join our Q&A platform to connect with experts dedicated to providing precise answers to your questions in different areas. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
Answer:
Approximately [tex]4.2\; {\rm s}[/tex] (assuming that the projectile was launched at angle of [tex]35^{\circ}[/tex] above the horizon.)
Explanation:
Initial vertical component of velocity:
[tex]\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing [tex]y_{1}[/tex] is the same as the altitude [tex]y_{0}[/tex] at which this projectile was launched: [tex]y_{0} = y_{1}[/tex].
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is [tex]20.6\; {\rm m\cdot s^{-1}}[/tex] (upwards,) the vertical velocity right before landing would be [tex](-20.6\; {\rm m\cdot s^{-1}})[/tex] (downwards.) The change in vertical velocity is:
[tex]\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Since there is no drag on this projectile, the vertical acceleration of this projectile would be [tex]g[/tex]. In other words, [tex]a = g = -9.81\; {\rm m\cdot s^{-2}}[/tex].
Hence, the time it takes to achieve a (vertical) velocity change of [tex]\Delta v_{y}[/tex] would be:
[tex]\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}[/tex].
Hence, this projectile would be in the air for approximately [tex]4.2\; {\rm s}[/tex].
Answer:
4.21 s
Explanation:
Vertical component of velocity = 36 sin 35 = 20.649 m/s
Vertical position is given by
yf = y0 + vo t - 1/2 at^2 yf = yo = 0 (ground level)
0 = 0 + 20.649 m/s * t - 1/2(9.81)t^2
t ( 20.649 - 4.905 t) = 0 show t = 0 and 4.21 s
the t = 0 is launch 4.21 seconds is landing
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.
