Playing jackstone as a kid, you discovered that there is a certain time before the rubber ball falls and you took advantage of this when picking up the stars on the floor. If you threw the ball 4.10m/s up into the air, how fast will it be moving at a height of 42.45cm? How much time has elapsed upon reaching this height?

Question

Playing jackstone as a kid, you discovered that there is a certain time before the rubber ball falls and you took advantage of this when picking up the stars on the floor. If you threw the ball 4.10m/s up into the air, how fast will it be moving at a height of 42.45cm? How much time has elapsed upon reaching this height?

Answer 1

To determine the speed of the ball at a height of 42.45 cm, we can use the principles of projectile motion and the laws of motion. We'll need to break this problem into two parts: finding the time it takes to reach the given height and then calculating the speed at that height.

Step 1: Finding the time (t) taken to reach the given height:
We can use the kinematic equation for vertical motion:

```
h = (v_iy * t) + (0.5 * a * t^2)
```
where:
h = height (42.45 cm = 0.4245 m)
v_iy = initial vertical velocity (throwing up means it's +4.10 m/s)
a = acceleration due to gravity (-9.8 m/s^2)

Rearranging the equation for time:
```
t = (-v_iy ± sqrt(v_iy^2 - 2 * a * h)) / a
```
Plugging in the given values:
```
t = (-4.10 ± sqrt(4.10^2 - 2 * (-9.8) * 0.4245)) / -9.8
```
Using the quadratic formula, we can solve for t.

Step 2: Calculating the speed (v) at the given height:
Once we have the time (t), we can calculate the velocity using the following equation:

```
v = v_iy + (a * t)
```
where:
v_iy = initial vertical velocity (4.10 m/s)
a = acceleration due to gravity (-9.8 m/s^2)

Now let's solve these equations to find the answers.

Note: We will consider the positive root for time (t) since it represents the upward motion.

Step 1: Calculating time (t):

```
t = (-(-4.10) + sqrt((-4.10)^2 - 2 * (-9.8) * 0.4245)) / -9.8
t = (4.10 + sqrt(16.81 + 8.3036)) / -9.8
t = (4.10 + sqrt(25.1136)) / -9.8
t = (4.10 + 5.01) / -9.8
t = 9.11 / -9.8
t = -0.9296 seconds
```

Step 2: Calculating velocity (v):

```
v = 4.10 + (-9.8 * -0.9296)
v = 4.10 + 9.11248
v ≈ 13.2125 m/s
```

Therefore, the speed of the ball at a height of 42.45 cm is approximately 13.2125 m/s. The time elapsed upon reaching this height would be approximately 0.9296 seconds.