Q#1: A rock dropped vertically from a cliff hits the ground in 4.00s (a) How hig

Question

Q#1: A rock dropped vertically from a cliff hits the ground in 4.00s

(a) How high is the cliff?

(b) What its speed just before impact?

(c) How far will it travel between t = 1.00 and t = 2.00 s?

Q#2: A certain car can reach 60mph in 9.8 s, starting from rest.

(a) What is the average accelaeration of the car, in ft/s^2?

(b) Assume the car is treveling 65mph. If It can decelerate at 3.5 ft/s^2 on dry pavement and 2 ft/s^2 on wet pavement. How much more distance is needed in order to stop from 65 mi/h on wet pevement than dry?

Posted by Bilawal

Explanation / Answer

a) h = ut+1/2at^2

intial velocity u = 0

h = 1/2*9.8*4^2 = 78.4 m

b) v = u+at = 9.8*4 = 39.2 m/s

c) y = u+1/2at^2

at 1 sec

v1 = 9.8*1 = 9.8 m/s

at 2 sec

v2 = 9.8+9.8*1 = 19.6 m/s

at 1 sec , y1 = 1/2*9.8*1^2 = 4.9 m

at 2 sec, y2 = 1/2*9.8*2^2 = 19.6 m

y = y2-y1 = 14.7 m

2) 1 mph = 1.47 ft/s

u = 0

v = 60 mph = 88 ft/s

a avg = 88/9.8 = 8.98 ft/s^2

b) v^2-u^2 = 2as

s = (65*1.47)^2/2*2 = 2270.5 ft = 692.05 m

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