1.) CSI find that a car hit thr ground 60m from the point where it left the clif
ID: 586154 • Letter: 1
Question
1.) CSI find that a car hit thr ground 60m from the point where it left the cliff. The cliff is 45m high. What speed was the car going when it left the cliff?
2.) A speeding car moving at a constatn velcocity of 50 m/s passes a policeman who immediatelyt gets on his motorcycle in order to catch up with that ar. How much time (in seconds) will pass before this happens? Assume the motorcycle has a cosntant acceleation of 4m/s and intialy has no velocity.
3.) A rocket, initially at rest, is fired veritcally such that its net acceleration is 10m/s2 in the upward direction. At an altitude of .50 km, the engine of the rocket cuts off. What is the maximum altitude it acheives?
4.) A showball rolls off a barn roof which is inclined at an angle of 40o. It leaves the roof with a velocity of 7m/s2 The edge of the roof is 14 m above the ground. Approximately how far from the base of the builidng does the snow strike the ground.
Explanation / Answer
1)
along vertical
initial velocity voy = 0
acceleration ay = -g
displacement y = -h
from equation of motion
y = voy*t + 0.5*ay*t^2
h = 0.5*g*t^2
t = sqrt(2h/g)
along horizantal
x = vx*t
60 = v*sqrt(2*45/9.8)
v = 19.8 m/s
++++++++++++
2)
for motor cyclist
x = u*t
for police man
x = v*t + 0.5*a*t^2
ut = v*t + 0.5*a*t^2
(50*t) = (0)+(0.5*4*t^2)
t = 25 s <-asnwer
+++++++
3)
along vertical
vy^2 - voy^2 = 2*a*y
vy^2 - 0^2 = 2*10*500
vy = 100 m/s
maximium height H = y + vy^2/2g
H = 500 + 100^2/(2*10) = 1 km
++++++++++++++++
4)
along vertical
-h = -v*sin40*t - 0.5*g*t^2
-14 = -(7*sin40*t) - (0.5*9.8*t^2)
t = 1.29 s
x = vx*t
x = v*cos40*t = 7*cos40*1.29 = 7 m <<<--answer
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