RaySix

Post

Displacement of a Thrown Ball

Created by @nathanedwards

at November 1st 2023, 9:45:05 am.

AP_Physics_1-Questions

#motion

#displacement

#projectile-motion

Question:

A ball is thrown straight upward with an initial velocity of 15 m/s. The ball reaches its maximum height after 3 seconds and then falls back down. Calculate the displacement of the ball from its initial position after 5 seconds.

Answer:

To calculate the displacement of the ball after 5 seconds, we need to consider the ball's motion during different intervals of time.

Let's break down the time intervals as follows:

From 0 seconds to 3 seconds: The ball travels upwards, reaching its maximum height.
From 3 seconds to 5 seconds: The ball falls back down to its initial position.

First, let's calculate the maximum height the ball reaches during the upward journey. We can use the equation:

v_f = v_i + a \cdot t

Where:

$v_f$ is the final velocity, which is 0 m/s at the peak of the motion.
$v_i$ is the initial velocity, which is 15 m/s.
$a$ is the acceleration due to gravity, which is -9.8 m/s^2 (negative because it acts in the opposite direction).
$t$ is the time taken to reach the maximum height, which is 3 seconds.

Rearranging the equation, we have:

0 = 15 + (-9.8) \cdot 3

Simplifying, we get:

0 = 15 - 29.4

29.4 = 15

Therefore, the maximum height the ball reaches is 29.4 meters.

Next, let's calculate the displacement during the falling phase from 3 seconds to 5 seconds. Since the ball starts at the maximum height, its displacement will be the negative of the distance fallen.

We can use the equation of motion:

d = v_i \cdot t + \frac{1}{2} a \cdot t^2

Where:

$d$ is the displacement.
$v_i$ is the initial velocity, which is 0 m/s.
$a$ is the acceleration due to gravity, which is -9.8 m/s^2 (negative because it acts in the opposite direction).
$t$ is the time taken during this interval, which is 5 - 3 = 2 seconds.

Plugging in the values, we have:

d = 0 \cdot 2 + \frac{1}{2} \cdot (-9.8) \cdot (2)^2

Simplifying, we get:

d = -19.6

Therefore, the displacement during the downward phase is -19.6 meters.

Finally, to calculate the total displacement after 5 seconds, we sum up the displacement during the upward phase and the displacement during the downward phase:

\text{Total displacement} = \text{Maximum height} + \text{Displacement during the falling phase}

\text{Total displacement} = 29.4 + (-19.6)

\text{Total displacement} = 9.8

Therefore, the ball is displaced 9.8 meters from its initial position after 5 seconds.

Note: The negative sign in the displacement during the falling phase indicates that the ball is in the opposite direction of the initial motion.