Answer to Question #157033 in Classical Mechanics for diorama

Question

 A small is eating dinner on a flat table that is 1 meter tall. When asked to eat his vegetables, the child pushes his plate away from him at a speed of 4 m/s. Due to friction, the plate decelerates at a constant rate of 1 m/s2. After 1 second, the plate falls off the edge of the table.

a.     How far did the plate travel over the 1 second?

 

 

 

  

 

b.     What is the speed of the plate at the instant it falls off the table?

 

 

 

 

 

c.     How much time will pass between the instant the plate leaves the table and the instant it reaches the ground?

 

 

 

 

 

d.     At what speed will the plate strike the ground?

 

 

 

 

e.     At what angle will the plate strike the ground?

 

Answer 1

$\text{the movement of the plate on the table is uniformly accelerated}$

$\text{with negative acceleration}$

$V(t) = V_0 -at$

$S(t) =V_0t -\frac{at^2}{2}$

$\text{where }a =1 m/s^2;V_0=4m/s$

$a)S(1)=4*1-\frac{1*1}{2}=3.5m$

$b)\text{by condition, the plate reaches the edge of the table in 1 second}$

$V(1) = 4 - 1 = 3m/s$

$\text{when reaching the end of the table,}$

$\text{the plate begins to fall freely}$

$c)t= \sqrt\frac{2h}{g}= \sqrt\frac{2*1}{9.8}\approx0.45s$

$d)\text{vertical fall rate}$

$V_y=gt= 0.45*9.8\approx4.43m/s$

$V= \sqrt{V_x^2+V_y^2}=\sqrt{3^2+4.43^2}\approx5.35m/s$

$e)\text{angle of incidence}$

$\cos\alpha= \frac{V_x}{V}=\frac{3}{5.35}\approx0.56$

$\alpha\approx56^0$

Answer:a)3.5m b)3m/s c)0.45 s d) 4.43m/s e)$56^0$