- #1
sillybean
- 18
- 0
[SOLVED] How far does arrow penetrate into board
An arrow is fired with a speed of 21.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1560 m/s^2 and the block's acceleration has a magnitude of 450 m/s^2.
a)How long does it take for the arrow to stop moving with respect to the block?
b)What is the common speed of the arrow and block when this happens?
c)How far into the block does the arrow penetrate?
x=Xi+Vi*t+ 1/2at^2
x=0+21m/s (1.04x10^-2 s)+ 1/2 (-1560m/s^2)(1.04*10^-2 s)^2
x=0.2184-0.0843648
x=0.134m
I got parts a and b. But part c is giving me problem. I tried using the equation above to get the final position of the arrow but the answer was wrong. Am I using the wrong equation. Or is it a math error?
Homework Statement
An arrow is fired with a speed of 21.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1560 m/s^2 and the block's acceleration has a magnitude of 450 m/s^2.
a)How long does it take for the arrow to stop moving with respect to the block?
b)What is the common speed of the arrow and block when this happens?
c)How far into the block does the arrow penetrate?
Homework Equations
x=Xi+Vi*t+ 1/2at^2
x=0+21m/s (1.04x10^-2 s)+ 1/2 (-1560m/s^2)(1.04*10^-2 s)^2
x=0.2184-0.0843648
x=0.134m
The Attempt at a Solution
I got parts a and b. But part c is giving me problem. I tried using the equation above to get the final position of the arrow but the answer was wrong. Am I using the wrong equation. Or is it a math error?