Practice Problem 0066
The GBM is chasing DB on top of a kitchen counter. DB, unaware that he is near the edge of the
counter, turns around and says, “You can’t get me”. Suddenly, DB steps off the counter and plunges
towards the hard kitchen floor. He lands 0.777 m from the edge of the base of the counter top.
Assume he has no initial vertical component of motion and that his initial speed is 2 m/s (only in the x
direction). Solve for the height of the countertop.
2 m/s
Δdy = ?
0.777 m
Begin by listing the information for each direction
x
y
Δdx = 0.777 m
ay = -9.8 m/s2
Vx = 2 m/s
ax = 0
Velocity in the x direction is constant since we assume the only force acting on the projectile is
gravity. Gravity only causes changes to the projectile’s velocity in the y direction.
For this problem, note that as the projectile launches off the countertop, its velocity is in the x direction
only, hence the initial velocity in the y direction is 0 (V1y = 0)
x
y
Δdx = 0.777 m
ay = -9.8 m/s2
Vx = 2 m/s
V1y = 0
ax = 0
Δdy = ?
Unfortunately, to solve for the displacement in the y direction, 3 variables need to be given prior to
using the kinematic equations. Hence, we will turn to the x direction to help us, since we know 3
variables.
For the x direction, we can determine the time (from when the projectile launches to when it hits the
ground).
v=
Δd
Δt
0.777
Δt
0.777
Δt =
2
2=
Δt = 0.3885 seconds
This time can be used to complete the problem.
x
y
Δdx = 0.777 m
ay = -9.8 m/s2
Vx = 2 m/s
V1y = 0
ax = 0
Δt = 0.3885 seconds
Δt = 0.3885 seconds
Focusing on the information from the y direction
Δdy = V1yΔt + 0.5ayΔt2
Δdy = (0)(0.3885) + 0.5(9.8)(0.3885)2
Δdy = 0 + 4.9(0.3885)2
Δdy = 0.74 m