Question

Imagine that you throw a rock into the air. When you release the rock, it is moving at 19.4 m/s. The rock flies straight up into the air and returns back to your hand. For how long was the rock in the air, considering the acceleration due to gravity is -9.81 m/s²?

A) 3.94 seconds
B) 1.98 seconds
C) 4.97 seconds
D) 2.49 seconds

Answer 1

Using the kinematic equation where displacement is zero and solving for time, we find that the rock was in the air for approximately 3.95 seconds. As the options provided are rounded, the closest option is A) 3.94 seconds for the total time the rock was airborne.

Step-by-step explanation:

To calculate how long the rock was in the air, we can use the kinematic equation for uniformly accelerated motion without initial displacement:

s = ut + \frac{1}{2}at^2

In this case, the displacement s is zero since the rock returns to the thrower's hand, the initial velocity u is 19.4 m/s, and the acceleration a due to gravity is -9.81 m/s². Setting displacement to zero and solving the quadratic equation for time t gives us the total time the rock spends in the air.

0 = (19.4 m/s)t + (1/2)(-9.81 m/s²)t²

This simplifies to:

0 = t(19.4 - 4.905t)

From this equation, one solution for t is zero (at the moment the rock is thrown), and the other is when the t value inside the parenthesis equals zero. Thus, we find the time by:

19.4 m/s = 4.905t

t = 19.4 / 4.905 = 3.95 seconds (approximately)

Since the total time includes the ascent and descent of the rock, and the time to reach the maximum height is half of the total time, the correct answer is A) 3.94 seconds, which is roughly half of 3.95 seconds (rounding down slightly since the options provided are not to two decimal places).