Question

An arrow is shot straight up in the air with an initial speed of 250 ft/s. If on striking the ground, it embeds itself 4.00 in into the ground, find the magnitude of the acceleration (assumed constant) required to stop the arrow in units of ft/sec^2

Answer 1

93750 ft/s²

Step-by-step explanation:

t = Time taken

u = Initial velocity = 250 ft/s (It is assumed that it is speed of the arrow just when it enter the ground)

v = Final velocity = 0

s = Displacement = 4 in =
`(4)/(12)=(1)/(3)\ feet`

a = Acceleration

Equation of motion

`v^2-u^2=2as\\\Rightarrow a=(v^2-u^2)/(2s)\\\Rightarrow a=(0^2-250^2)/(2* (1)/(3))\\\Rightarrow a=-93750\ ft/s^2`

The magnitude of acceleration is 93750 ft/s²