Question

1. A beanbag is thrown horizontally from a dorm room window. The height of the window is h = 10 meters above the ground as shown in the figure. If the beanbag

Answer 1

this way is a bit long, but it goes step by step and uses the usual kinematic equations

find the vertical velocity Vy before hitting ground, use equation: (v2)^2 = (v1)^2 + 2ad

for a vertical drop of 10 meters, v1=0, d = 10m, a = 9.81m/s^2
(v2)^2 = (v1)^2 + 2ad
(v2)^2 = 2ad
(v2) = sqrt(2ad)
v2 = sqrt(2*9.81*10)
v2 = 14 m/s
Vy = 14 m/s, this is the vertical velocity of the bag before hitting ground

find the time the bag flies which is the time it takes the bag to fall and reach Vy under that acceleration of gravity:
v = at
t = v/a
t = 14 / 9.81 seconds (leave it in this form)

find the magnitude of the beanbag velocity before hitting ground
Vsin? = Vy
Vsin(30deg) = 14
V = 14 / sin(30deg)
V = 28 m/s

find the x component (Vx) of the velocity V
Vx = Vcos?
Vx = 28*cos(30deg)
Vx = 28*cos(30deg) (leave it in this form)

find the distance using Vx and the total time the beanbag flies
distance = Vx * t
distance = 28 * cos(30deg) * (14 / 9.81)
distance = 34.6057042083

the beanbag travels a horizontal distance of about 34.6 m