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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
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