A cat chases a mouse across a 1.48 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. What was the cat’s speed when it slid off the table? At what velocity did the cat hit the floor?
The Correct Answer and Explanation is:
We can solve this problem using principles of projectile motion. Here’s a step-by-step approach:
1. Determine the time it takes for the cat to fall:
The cat falls vertically from a height of 1.48 m, so we can use the following kinematic equation to find the time it takes to hit the ground:y=12gt2y = \frac{1}{2} g t^2y=21gt2
Where:
- yyy is the height (1.48 m),
- ggg is the acceleration due to gravity (9.8 m/s²),
- ttt is the time in seconds.
Rearranging the equation to solve for time:t=2yg=2(1.48)9.8=2.969.8=0.302≈0.55 secondst = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2(1.48)}{9.8}} = \sqrt{\frac{2.96}{9.8}} = \sqrt{0.302} \approx 0.55 \, \text{seconds}t=g2y=9.82(1.48)=9.82.96=0.302≈0.55seconds
2. Determine the horizontal speed of the cat:
Since the horizontal motion is independent of the vertical motion, we can use the horizontal distance the cat travels (2.2 m) and the time it takes to fall (0.55 s) to find the cat’s initial horizontal velocity.
The horizontal velocity vxv_xvx is given by:vx=horizontal distancetime=2.2 m0.55 s=4.0 m/sv_x = \frac{\text{horizontal distance}}{\text{time}} = \frac{2.2 \, \text{m}}{0.55 \, \text{s}} = 4.0 \, \text{m/s}vx=timehorizontal distance=0.55s2.2m=4.0m/s
3. Determine the velocity when the cat hits the ground:
The cat’s final velocity will have both a horizontal and a vertical component. We already know the horizontal velocity (vx=4.0 m/sv_x = 4.0 \, \text{m/s}vx=4.0m/s).
To find the vertical velocity when the cat hits the ground, we can use the equation for the final velocity under constant acceleration:vy=gt=9.8 m/s2×0.55 s=5.39 m/sv_y = g t = 9.8 \, \text{m/s}^2 \times 0.55 \, \text{s} = 5.39 \, \text{m/s}vy=gt=9.8m/s2×0.55s=5.39m/s
4. Determine the total velocity:
The total velocity when the cat hits the ground is the vector sum of the horizontal and vertical velocities. We can calculate it using the Pythagorean theorem:v=vx2+vy2=(4.0)2+(5.39)2=16+29.05=45.05≈6.71 m/sv = \sqrt{v_x^2 + v_y^2} = \sqrt{(4.0)^2 + (5.39)^2} = \sqrt{16 + 29.05} = \sqrt{45.05} \approx 6.71 \, \text{m/s}v=vx2+vy2=(4.0)2+(5.39)2=16+29.05=45.05≈6.71m/s
Final Answers:
- The cat’s speed when it slid off the table was 4.0 m/s4.0 \, \text{m/s}4.0m/s.
- The velocity with which the cat hit the floor was approximately 6.71 m/s6.71 \, \text{m/s}6.71m/s.
