Question

Calvin tosses a water balloon to Hobbs. As Hobbs is about to catch the balloon it has a speed of 1 m/s. Hobbes catches the balloon, and a balloon experiences an acceleration of -0.5 m/s squared as it comes to rest. How far did Hobbes' hands move back while catching the balloon?

Answer 1

Hobbes' hands moved back 1 meter while catching the water balloon by using the kinematic equation v^2 = u^2 + 2as, where the initial speed was 1 m/s and the acceleration was -0.5 m/s^2.

Step-by-step explanation:

To determine how far Hobbes' hands move back while catching the balloon, we can use the kinematic equations of motion. Given that the final speed (v) of the balloon is 0 m/s (since it comes to rest), the initial speed (u) is 1 m/s, and the acceleration (a) is -0.5 m/s2, we can use the equation v2 = u2 + 2as to solve for the distance (s) moved by Hobbes' hands. Plugging in our values, we get:

0 = (1 m/s)2 + 2(-0.5 m/s2)s.

Solving for s, we find s = 1 m2/s2 / (2 * -0.5 m/s2) which simplifies to s = 1 m.

Therefore, Hobbes' hands moved back a distance of 1 meter while catching the balloon.