Question

A ball is dropped from the height of 10m.at the same time, another ball is thrown vertically upward at an initial speed of 10m/s. How high above the ground will thr two balls collide?

Answer 1

Given Information:

Ball One:

`\vec y_(0) = 10 \ m` (Indicating the initial position)

We also know the ball was dropped from rest. So,
`\vec v_{0_(1) } = 0 \ m/s`.

Ball Two:

`\vec v_(0) = 10 \ m/s` (Indicating the initial velocity)

We also know the ball was throw from the ground. So,
`\vec y_{0_(1) } = 0 \ m`.

The Information we want to Find:

`\vec y_(c) = ?? \ m` (Indicating the position the two projectiles collide)

Using the Following Kinematic Equation to Solve:

`\Delta \vec x = \vec v_(0)t + (1)/(2) \vec at`

For ball one...

`\Delta \vec y = \vec v_(0)t + (1)/(2) \vec a_(y) t`

`\Longrightarrow \vec y_(c)- \vec y_(0) = \vec v_{0_(1) }t + (1)/(2) \vec a_(y)t`

`\Longrightarrow \vec y_(c)- \vec y_(0) = (0)t + (1)/(2) \vec a_(y)t`

`\Longrightarrow \vec y_(c)- \vec y_(0) = (1)/(2) \vec a_(y)t`

`\Longrightarrow \vec y_(c) = (1)/(2) \vec a_(y)t +\vec y_(0)` => Equation 1

For ball two...

`\Delta \vec y = \vec v_(0)t + (1)/(2) \vec a_(y)t`

`\Longrightarrow \vec y_(c)- \vec y_{0_(1) } = \vec v_(0)t + (1)/(2) \vec a_(y)t`

`\Longrightarrow \vec y_(c) = \vec v_(0)t + (1)/(2) \vec a_(y)t +\vec y_{0_(1) }`

`\Longrightarrow \vec y_(c) = \vec v_(0)t + (1)/(2) \vec a_(y)t + 0`

`\Longrightarrow \vec y_(c) = \vec v_(0)t + (1)/(2) \vec a_(y)t` => Equation 2

Set equations 1 and 2 equal to each other and solve for the time that they collide.

`\left \{ {{\vec y_(c) = (1)/(2) \vec a_(y)t +\vec y_(0)} \atop { \vec y_(c) = \vec v_(0)t + (1)/(2) \vec a_(y)t } \right.`

`\Longrightarrow (1)/(2) \vec at +\vec y_(0)= \vec v_(0)t + (1)/(2) \vec at`

`\Longrightarrow \vec y_(0)= \vec v_(0)t`

`\Longrightarrow t=(\vec y_(0))/(\vec v_(0))`

`\Longrightarrow t=(10)/(10)`

`\Longrightarrow t=1 \ s`

Thus, the balls collide at time, t=1 s. We can now use this time to plug into equation 1 or 2 to find the height at which they collide. I will use equation 1.

`\Longrightarrow \vec y_(c) = (1)/(2) \vec a_(y)t +\vec y_(0)`

`\Longrightarrow \vec y_(c) = (1)/(2) (-9.8)(1) +10`

`\Longrightarrow \vec y_(c) = 5.1 \ m \ \therefore \ Sol.`

*Note*
`\vec a_(y)` is the acceleration of gravity (
`-9.8 \ m/s^2 \ or \ -32 \ ft/s^2`)

Final Answer: The balls collide at the height 5.1 m.

Answer 2

Answer: H=5.1m

Step-by-step explanation:

Given:

Ball 1 height= 10m

Ball 2 initial velocity=10m/s

use the kinematic equation:

S=(vi)t+12at2

I choose my sign convention to be up=positive, down=negative, so a=−9.81ms2 (a is taken as the value of gravity)

Ball 1 dropped from 10m :

−(10−H)=0+12(−9.81)t2

Note that (10-S) is negative because that displacement is *below* the starting point.

12(9.81)t2=10−H

——- equation (1)

Ball 2 thrown upward at 10 m/s :

H=(10)t+12(−9.81)t2

or

12(9.81)t2=10t−H

——- equation (2)

equation (1) minus equation (2):

0=(10−H)−(10t−H)

t=1 equation (1):

12(9.81)12=10−H

H=5.1m