Say the horizontal component of the velocity is vx and the vertical is vy.
Initially at t=0 (as the mug leaves the bar) the components are v0x and v0y.
Obviously (I hope!) v0y = 0.
The equations for horizontal and vertical projectile motion (with the positive direction up) are
x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2
Now choose the origin to be the end of the counter. x0=0 and y0=0. The equations simplify to
x = v0x t
y = - 1/2 g t^2
You know that x = 1.20m when y = -0.88m
From the y equation (and g=10 m/s^2) you can calculate the time that the mug hits the floor.
t = 0.420s
From the x equation we get the initial horizontal velocity
v0x = x/t = 1.2/0.42 = 2.86 m/s
(b) x-component of velocity is constant since there are no horizontal forces so vx = 2.86 m/s
y-component is given by v = u+at with u=0 and a=-g
vy = -gt = -4.2m/s
Now tan(angle) = vy/vx so angle = arctan(vy/vx)
If its uniform there is no force.
C) 0 N
From your choices which are
A. 5.6 × 10-12 joules
B. 1.3 × 10-12 joules
C. 2.3 × 10-12 joules
D. 3.3 × 10-10 joules
E. 5.6 × 10-10 joules
The answer is 3.3 × 10-10 joules. The answer is letter c.
Answer:
Explanation:
We are going to use the accelerated motion formula 
on the vertical component taking the upwards direction as positive.
Since <em>at maximum height the speed will be 
</em>, the formula can be written as 
, which for our values is:
Written to three decimal places.
<span>Vf^2=Vi^2+2ad
you know Vf, Vi, mg, and distance d.
solve for frictionforceResponding to an alarm, a 100 kg, a fireman slides down a pole to the ground floor, 3.1 m below. The fireman starts at rest and lands with a speed of 3.8 m/s. Find the average force exerted on the fireman by the pole. Find the average force exerted on the fireman when the landing speed is 1.9 m/s.</span>
