Question

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?

Answer 1

GO THROUGH THE ATTACHED IMAGE ❣️....

please mark me BRAINIEST ♥️..

Answer 2

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!!!

Answer 3

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!!!