An Archer shoots an arrow horizontally at 250 feet per second. The bullseye on the target and the arrow are initially at the same height. If the target is 60 feet from the archer, how far below the bullseye (in feet) will the arrow hit the target? Neglect any effects of air resistance.(Note: g=9.8m/s2 = 32 feet/s2)

d = ft

0.92 ft.

Explanation:

Equations of motion: v = distance/time

i. vf = vi + a*t

ii. S = vi*t + 1/2*(a*(t^2))

t = time taken to reach target

S = distance to target

vf = final velocity of arrow

vi = initial velocity

Given:

vf = 250 ft/s

S = 60 ft

t = 60/250

t = 0.24 s.

Using ii. equation,

distance fallen by arrow, S = 1/2 * 32 * (0.24^2)

= 0.92 ft.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

a displacement distance is size of displ