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Mekhanik [1.2K]
3 years ago
15

A child’s toy arrow is launched horizontally from a bow at a speed of 8 m/s from a second story window 4 meters above the ground

, and the sticks in a pile of sand.
A. How far away is the pile of sand?

B. What were the horizontal and vertical speeds of the arrow when it hits the sand?

C. What angle does the arrow make with the ground when it sticks in the sand?
Physics
1 answer:
Westkost [7]3 years ago
7 0
First find the time it takes to hit the sand t
using s=1/2 g t^2
t^2= 8/9.8= 0.82 seconds
This means the sand was s = 8 * 0.82 = 6.5m away
The horizontal speed stays constant at 8 m/s
The vertical speed =.82 * 9.8 = 8.0 m/s
As the two velocities are equal the angle must be 45 degrees to the horizontal
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Answer:

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T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

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\Delta x = 9.7*10^{-3}m

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T=\frac{Ed}{Rln(\frac{J\Delta x}{D_0 \Delta c})}

T=-\frac{-82000}{(8.31)ln(\frac{3.2*10^{-9}(9.7*10^{-3})}{6.5*10^{-7} (0.325)})}

T=1118.07K=844\°C

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