A. Let’s
say the horizontal component of the velocity is vx and the vertical is vy.
Initially at t=0 (as the mug leaves the counter) the
components are v0x and v0y.
v0y = 0 since the customer slides it horizontally so applied
force is in the x component only.
The equations for horizontal and vertical projectile motion
are:
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2
Setting the origin to be the end corner of the counter so that x0=0 and y0=0, hence:
x = v0x t
y = - 1/2 g t^2
Given value are: x=1.50m and y=-1.15m (y is negative since mug is going down)
1.50m = v0x t ----> v0x= 1.50/t
-1.15m = -(1/2) (9.81) t^2 -----> t =0.4842 s
Calculating for v0x:
v0x = 3.10 m/s
B. v0x is constant since there are no other horizontal forces so, v0x=vx=3.10m/s
vy can be calculated from the formula:
vy = v0y + at where a=-g (negative since going down)
vy = -gt = -9.81 (0.4842)
vy = -4.75 m/s
Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy
tan(90-Ø )= 3.1/4.75
Ø = 56.87˚ below the horizontal
