Question

The cat jumped 2.5 meters from the 2-meter-high fence and landed 2.8 meters in horizontal distance. What is the speed and angle at which the cat leaves the end of the fence?

Answer 1

Expression of max height

h = ( v sin x) ^2 / 2g

v sin x = sqrt ( 2.5* 2* 9.8)

v sin x = 7.... (i)

Motion along vertical

h = ho + v sin x * t - 0.5 gt^2

0 = 2 + 7 t - 4.9 t^2

Solving above quadratic equation

t = 1.6726 s

Motion along horizontal

v cos x = 2.8 / t

v cos x = 2.8/1.6726

v cos x = 1.674 ... (ii)

Squaring and adding... (i) and.. (ii)

v^2 ( sin^2 x + cos^2 x) = 7^2 + 1.674^2

v = 7.197 m/s

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Putting in .... (i)

x = arcsin ( 7 / 7.197 )

x = 76.55 deg

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