A physics student is driving his pick-up truck down Lake Avenue. The pick-up is
equipped with a projectile launcher that
imparts a vertical velocity to a water-filled
rubber projectile. While traveling 20.0 m/s
in an eastward direction, the projectile is
launched vertically with a velocity of 58.8
m/s.
Fill in the table at the right. showing the
horizontal and vertical displacement of the
projectile every second for the first 12
seconds.

TOTAL FLIGHT TIME

Rising Projectile

vi = initial velocity = 58.8 m/s

vf = final velocity = 0

g = gravity acceleration = -9.81 m/s²

t = elapsed time = to be determined

vf = vi + gt

vf - vi = gt

(vf - vi)/g = t

t = (vf - vi)/g

t = (0 - 58.8)/-9.81

t = -58.0/-9.81

t = 5.99 s ≈ 6.00 s

Falling Projectile

vi = initial velocity = 0

vf = final velocity = 58.8 m/s

g = gravity acceleration = 9.81 m/s²

t = elapsed time = to be determined

vf = vi + gt

vf - vi = gt

(vf - vi)/g = t

t = (vf - vi)/g

t = (58.8 - 0)/9.81

t = 58.0/9.81

t = 5.99 s ≈ 6.00 s

t(total) = t(up) + t(down) = 6.00 s + 6.00 s = 12.0 s Total Flight Time

HORIZONTAL DISTANCE TRAVELED

v = velocity, horizontal = 20.0 m/s

d = to be determined

t = elapsed time = 12.0 s

v = d/t

vt = d

d = vt

d = (20.0 m/s)(12.0 s)

d = 240 m Horizontal Distance Traveled

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think its the third option

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