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sertanlavr [38]
3 years ago
6

If the window is 11 m above the ground, find the time the stone is in flight.

Physics
1 answer:
rjkz [21]3 years ago
5 0
d=vi*t+(1/2)gt²

d=11 m
g=9.8 m/s²
vi=0 m/s

11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s

Answer: the time the sone is in flight is 1.5 s
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In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope
zysi [14]

Answer:

With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.

Explanation:

Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude g.

If we have the initial velocity v_o and its angle \theta, we can obtain the vertical component of the velocity v_{oy} using trigonometry:

v_{oy}=v_osin\theta

Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:

v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }

Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:

g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

Finally, from the equation of horizontal motion with constant speed, we have that:

x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}

For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:

v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s}  }{9.8\frac{m}{s^{2} } } =7.49m

In words, the projectile travels 7.49m horizontally before it lands.

8 0
4 years ago
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