Question

A CAR MOVin0·AT /20 kph APPROACHES A VERTICAL CLIFF. THe DA, V6R see s тна CLIFF WHE" H6's qo.om FRom ,rs ebee ANO How FAR SHOAT R THE CLIFF DOES THE CAR SToP FisD He P^ RT.c.Les veLoS!TY AT t ㅇ. 7.9 mis. b How HIGH IS THE BALL AT ITS HIGHEST PDNT ④ IN PROP#A METRIC.UNITS、THE P0.5/rie N·ξ ㅅ PARTICLE ON TH E a FIND THE POSIT ION), VELOCITY, AND ACCELEAATION OF THE d) FIND THE POSITION OF THe PARTICLE WHEN MOMENTARILY AT REST. t(s) l6

Answer 1

(1)
Initial speed of car (u) = 120 km/h = 33.33 m/s
acceleration (a) = - 6.9 m/s²
Now using the kinematic equation distance covered by the car till it stops
V² = u² +2aS
0 = 33.33² +2*(-6.9)S
S = 80.5 m
Now the distance remain from the cliff = 90- 80.5 = 9.5 m
Hence the car will short by 9.5 m from the cliff.
(2)
Speed at t = 0
V = 62 m/s
acceleration of particle (a) = -2.3 m/s²
Now the velocity at t= 0
Using kinematic equation
V = u +at
62 = u +(-2.3)*(8-0)
u = 80.4 m/s
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