IDNLearn.com makes it easy to get reliable answers from experts and enthusiasts alike. Discover reliable answers to your questions with our extensive database of expert knowledge.
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
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.
