Ok so this is simple projectile motion problem.

if we have an object falling in free fall it is subject to gravity of -9.80m/s^2

so it says it takes 6 sec to fall and we know initial velocity was zero so we know that h=vt+1/2gt^2 so we get h=0+1/2*9.80*6^2 = 176.4m

so solving for final speed we get KE=PE = 1/2mv^2=mgh = 1/2v^2=gh so
v=sqrt(2*g*h) = sqrt(2*9.8*176.4m) = 58.8m/s final speed when it hits the ground

hope this helps you! Thanks!!

8 0

3 years ago

The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t

frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

$Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%$