Question

1) Sitting in a second-story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.24 s as it moves a distance of 1.08 m from the bottom to the top of the window.How long does it take before the ball reappears? (b)What is the greatest height of the ball above the top of the window?

2)To be compliant with regulations the inclination angle of a wheelchair ramp must not exceed 4.76∘.If the wheelchair ramp must have a vertical height of 1.11 m, what is its minimum horizontal length?

Answer 1

velocity at the bottom of window = v1

velocity at the top of window = v2

y = v1*t + (1/2)*ay*t^2

1.08 = v1*0.24 - (1/2)*9.8*0.24^2

v1 = 5.676 m/s

y = (v1+v2)/2*t

1.08 = (5.676+v2)*0.24/2

v2 = 3.324 m/s

the ball will reappear after time T = 2*v2/g

T = 2*3.324/9.8 = 0.0.68 s

(b)

maximum height H = v2^2/2g = 3.324^2/(2*9.8) = 0.564 m

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(2)

tantheta = h/L

L = h/tantheta

L = 1.11/tan4.76

L = 13.3 m