(a) h = 4.9t^2
(b) the horizontal speed is constant: 1.20 m/s, and distance = speed * time
(c) vx = 1.20 and vy = -gt
magnitude = √(vx^2 + vy^2)
tanθ = vy/vx
A Physics textbook slides off a horizontal tabletop with a speed of 1.20 m/s. It
strikes the floor in 0.50 s. Ignore air resistance. Find
-The height of the tabletop above the floor.
- the horizontal distance from the edge of the table to the point where the book strikes the
floor.
-The horizontal and vertical components of the book’s velocity, and the magnitude and
direction of its velocity just before the book reaches the floor.
2 answers
Divide these problems into a horizontal problem and a vertical problem.
They are connected by the time.
There is no horizontal force on the system. Therefore the horizontal momentum of your object is constant. If the mass is constant, the horizontal velocity component is constant
u = 1.20 m/s from the table to the floor.
in 0.5 seconds it lands 0.60 meters from the table
There IS a vertical force downward , m g
so if vertical velocity up is Vi at start and height is Hi then
v = Vi - g t
h = Hi + Vi t - (1/2) g t^2
at floor h = 0 so
0 = Hi + 0 - 4.9 t^2
they told us t = 0.50 s
so
Hi = 4.9 (0.5)^2 = 1.225 meters high
then v = g t = -9.81 * 0.5 m/s= - 4.9 m/s
u is still 1.20
so speed = sqrt (1.2^2 + 4.9^2) etc
They are connected by the time.
There is no horizontal force on the system. Therefore the horizontal momentum of your object is constant. If the mass is constant, the horizontal velocity component is constant
u = 1.20 m/s from the table to the floor.
in 0.5 seconds it lands 0.60 meters from the table
There IS a vertical force downward , m g
so if vertical velocity up is Vi at start and height is Hi then
v = Vi - g t
h = Hi + Vi t - (1/2) g t^2
at floor h = 0 so
0 = Hi + 0 - 4.9 t^2
they told us t = 0.50 s
so
Hi = 4.9 (0.5)^2 = 1.225 meters high
then v = g t = -9.81 * 0.5 m/s= - 4.9 m/s
u is still 1.20
so speed = sqrt (1.2^2 + 4.9^2) etc
2 answers