The formula to solve this is (.80)^3 X 6 and the answer would be 3.1 feet. That is how high the ball will rebound after its third bounce. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
I think the question should be the below:
<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:
<span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>
<span>x (max) = a(max) /ω² </span>
<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
By calculating the crests, you can find the waves' frequency.
Hope this helps!
v = initial velocity of launch of the stone = 12 m/s
θ = angle of the velocity from the horizontal = 30
Consider the motion of the stone along the vertical direction taking upward direction as positive and down direction as negative.
v₀ = initial velocity along vertical direction = v Sinθ = 12 Sin30 = 6 m/s
a = acceleration of the stone = - 9.8 m/s²
t = time of travel = 4.8 s
Y = vertical displacement of stone = vertical height of the cliff = ?
using the kinematics equation
Y = v₀ t + (0.5) a t²
inserting the values
Y = 6 (4.8) + (0.5) (- 9.8) (4.8)²
Y = - 84.1 m
hence the height of the cliff comes out to be 84.1 m
Biosphere is the answer, hope i helped :))
