Question

A water balloon was dropped from a high window and struck its target 1.17 seconds later. If the balloon left the person's hand at -4.5 m/sec, what was its velocity on impact?

Answer 1

Answer on Question 66724, Physics, Atomic and Nuclear Physics

Question:

A water balloon was dropped from a high window and struck its target 1.17 seconds later. If the balloon left the person's hand at $-4.5\ m/s$, what was its velocity on impact?

Solution:

We can find the velocity of the water balloon on impact from the kinematic equation:

$$v = v_i + g t,$$

here, $v_i = -4.5\ m/s$ is the initial velocity of the water balloon, $v$ is the velocity of the water balloon on impact, $g = -9.8\ m/s^2$ is the acceleration due to gravity and $t$ is the time.

Then, we get:

$$v = v_i + g t = -4.5 \frac{m}{s} + \left(-9.8 \frac{m}{s^2}\right) \cdot 1.17\ s = -15.96 \frac{m}{s}.$$

The sign minus indicates that the velocity of the balloon is directed downward.

Answer:

$$v = -15.96 \frac{m}{s}.$$

Answer provided by AssignmentExpert.com