Question

A started cat jumps off of the edge of a $1.00m$ high table with an initial velocity of $2.00m/s$ ${45}^o$ above the horizontal. How far away horizontally on the floor will the cat land relative to the table?
Take air resistance to be negligible and gravity to be the only source of acceleration:

• A. $20.4cm$
• B. $61.8m$
• C. $87.4cm$
• D. $139cm$
• E. Not enough information

Answer 1

Answer: (C)

$87.4cm$

The vertical component of velocity of cat=$+2m/s\times sin{45}^{\circ }$

$=+\sqrt{2}m/s$

From equation of motion,

$s=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$⟹-1=\sqrt{2}\left(t\right)-\frac{1}{2}\times 10t^2$

$⟹t=0.61s$

Thus horizontal distance=$ucos{45}^{\circ }\times t$

$=87.4cm$