Question

Physics help please

Ball is thrown horizontally from top of building 120 m high. Ball strikes ground 52 m horizontally from the point of release. What is the speed of the ball just before it strikes the ground? Answer in units of m/s

Answer 1

The speed of the ball just before it strikes the ground is approximately
`\( 29.4 \, \text{m/s} \).`

To find the speed of the ball just before it strikes the ground, we can use the following kinematic equation:

`\[ s = ut + (1)/(2)gt^2 \]`

where:

-
`\( s \)` is the vertical displacement (height of the building),

-
`\( u \)` is the initial vertical velocity (which is 0 since the ball is thrown horizontally),

-
`\( g \)` is the acceleration due to gravity, and

-
`\( t \)` is the time of flight.

The horizontal displacement
`(\( d \))` is given as 52 m. In horizontal motion, the velocity is constant, and we can use the formula:

`\[ d = vt \]`

where:

-
`\( v \)` is the horizontal velocity, and

-
`\( t \)` is the time of flight.

Since the ball is thrown horizontally, the initial vertical velocity
`(\( u \))` is 0, and the vertical displacement
`(\( s \))` is the height of the building (120 m). Rearrange the first equation to solve for
`\( t \)`:

`\[ t = \sqrt{(2s)/(g)} \]`

Now, substitute this time value into the second equation:

`\[ d = v \sqrt{(2s)/(g)} \]`

Solve for
`\( v \)`:

`\[ v = \frac{d}{\sqrt{(2s)/(g)}} \]`

Given
`\( s = 120 \, \text{m} \) and \( d = 52 \, \text{m} \)`, and
`\( g \approx 9.8 \, \text{m/s}^2 \)`, plug in these values to find
`\( v \)`:

`\[ v \approx \frac{52}{\sqrt{(2 * 120)/(9.8)}} \approx 29.4 \, \text{m/s} \]`

Answer 2

Answer: 37.981 m/s

Step-by-step explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

`x=V_(o)cos\theta t` (1)

Where:

`x=52 m` is the point where the ball strikes ground horizontally

`V_(o)` is the ball's initial speed

`\theta=0` because we are told the ball is thrown horizontally

`t` is the time since the ball is thrown until it hits the ground

y-component:

`y=y_(o)+V_(o)sin\theta t+(gt^(2))/(2)` (2)

Where:

`y_(o)=120m` is the initial height of the ball

`y=0` is the final height of the ball (when it finally hits the ground)

`g=-9.8m/s^(2)` is the acceleration due gravity

Knowing this, let's start by finding
`t` from (2):

`0=y_(o)+V_(o)sin(0\°) t+(gt^(2))/(2)` (3)

`0=y_(o)+(gt^(2))/(2)`

`t=\sqrt{(-2 y_(o))/(g)}` (4)

`t=\sqrt{(-2 (120 m))/(-9.8m/s^(2))}` (5)

`t=4.948 s` (6)

Then, we have to substitute (6) in (1):

`x=V_(o)cos(0\°) t` (7)

And find
`V_(o)`:

`V_(o)=(x)/(t)` (8)

`V_(o)=(52 m)/(4.948 s)` (9)

`V_(o)=10.509 m/s` (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity
`V`:

`V=V_(o) + gt` (11)

`V=10.509 m/s + (-9.8 m/s^(2))(4.948 s)` (12)

`V=-37.981 m/s` (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).