A cat jumps off a table which is 1.5 m high. If the initial velocity of the cat is 10 m/s, at an
angle of 37º above the horizontal, how far from the edge of the table does the cat land?
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The vertical component of velocity of cat=+2m/s×sin45 ∘=+ 2 m/s
From equation of motion,s=u y t+ 1
a y
t 2
⟹−1= 2
(t)− 21 ×10t 2
⟹t=0.61s
Thus horizontal distance=ucos45∘ ×t
=87.4cm
Brainliest plz!!!
Answered by
0
The vertical component of velocity of cat=+2m/s×sin45 ∘=+ 2 m/s
From equation of motion,s=u y t+ 1
a y
t 2
⟹−1= 2
(t)− 21 ×10t 2
⟹t=0.61s
Thus horizontal distance=ucos45∘ ×t
=87.4cm
Brainliest plz!!!
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