Answer :
Certainly! Let's solve the problem step by step for each part.
### Part (a): In what time, will it reach the maximum height?
To determine the time it takes for the ball to reach maximum height, we start by using the kinematic equation for velocity:
[tex]\[ v = u + at \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity (0 m/s at the maximum height),
- [tex]\( u \)[/tex] is the initial velocity (18 m/s),
- [tex]\( a \)[/tex] is the acceleration due to gravity (-9.8 m/s²), and
- [tex]\( t \)[/tex] is the time.
Setting [tex]\( v = 0 \)[/tex] at the maximum height, we get:
[tex]\[ 0 = 18 - 9.8t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{18}{9.8} \][/tex]
[tex]\[ t \approx 1.837 \, \text{seconds} \][/tex]
Therefore, the time to reach the maximum height is approximately 1.837 seconds.
### Part (b): How high above the building will the ball rise?
To find the maximum height reached above the building, we use the kinematic equation for displacement:
[tex]\[ h = ut + \frac{1}{2}at^2 \][/tex]
Where:
- [tex]\( h \)[/tex] is the maximum height above the building,
- [tex]\( u \)[/tex] is the initial velocity (18 m/s),
- [tex]\( t \)[/tex] is the time to reach maximum height (1.837 seconds),
- [tex]\( a \)[/tex] is the acceleration due to gravity (-9.8 m/s²).
Substituting the values:
[tex]\[ h = 18 \times 1.837 + \frac{1}{2} \times (-9.8) \times (1.837)^2 \][/tex]
[tex]\[ h \approx 33.066 - 16.535 \][/tex]
[tex]\[ h \approx 16.531 \, \text{meters} \][/tex]
Thus, the ball will rise approximately 16.531 meters above the building.
### Part (c): Compute the velocity with which it will strike the ground and the total time it is in motion
1. Velocity with which it will strike the ground:
First, we determine the total height the ball will fall from, which is the initial height of the building plus the maximum height above the building:
[tex]\[ \text{Total height fall} = 25 \, \text{meters} + 16.531 \, \text{meters} \][/tex]
[tex]\[ \text{Total height fall} = 41.531 \, \text{meters} \][/tex]
Using the kinematic equation for velocity where the fall starts from rest:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( u \)[/tex] is the initial velocity (0 m/s from the maximum height),
- [tex]\( a \)[/tex] is the acceleration due to gravity (9.8 m/s²),
- [tex]\( s \)[/tex] is the total height fall (41.531 meters).
Substituting the values:
[tex]\[ v^2 = 0 + 2 \times 9.8 \times 41.531 \][/tex]
[tex]\[ v^2 = 812.414 \][/tex]
[tex]\[ v \approx \sqrt{812.414} \][/tex]
[tex]\[ v \approx 28.531 \, \text{m/s} \][/tex]
Therefore, the velocity with which the ball will strike the ground is approximately 28.531 m/s.
2. Total time in motion:
The total time of motion includes the time to reach the maximum height and the time to fall back to the ground. We already have the time to reach maximum height as 1.837 seconds.
To find the time to fall from the maximum height to the ground, we use the kinematic equation for displacement:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Where:
- [tex]\( s \)[/tex] is the total height fall (41.531 meters),
- [tex]\( u \)[/tex] is the initial velocity (0 m/s from rest),
- [tex]\( a \)[/tex] is the acceleration due to gravity (9.8 m/s²).
Solving for [tex]\( t \)[/tex]:
[tex]\[ 41.531 = \frac{1}{2} \times 9.8 \times t^2 \][/tex]
[tex]\[ 41.531 = 4.9 \times t^2 \][/tex]
[tex]\[ t^2 \approx \frac{41.531}{4.9} \][/tex]
[tex]\[ t^2 \approx 8.478 \][/tex]
[tex]\[ t \approx \sqrt{8.478} \][/tex]
[tex]\[ t \approx 2.911 \, \text{seconds} \][/tex]
Therefore, the total time in motion is the sum of the time to reach maximum height and the time to fall back down:
[tex]\[ \text{Total time in motion} = 1.837 + 2.911 \][/tex]
[tex]\[ \text{Total time in motion} \approx 4.748 \, \text{seconds} \][/tex]
Hence, the total time it is in motion is approximately 4.748 seconds.
### Part (a): In what time, will it reach the maximum height?
To determine the time it takes for the ball to reach maximum height, we start by using the kinematic equation for velocity:
[tex]\[ v = u + at \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity (0 m/s at the maximum height),
- [tex]\( u \)[/tex] is the initial velocity (18 m/s),
- [tex]\( a \)[/tex] is the acceleration due to gravity (-9.8 m/s²), and
- [tex]\( t \)[/tex] is the time.
Setting [tex]\( v = 0 \)[/tex] at the maximum height, we get:
[tex]\[ 0 = 18 - 9.8t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{18}{9.8} \][/tex]
[tex]\[ t \approx 1.837 \, \text{seconds} \][/tex]
Therefore, the time to reach the maximum height is approximately 1.837 seconds.
### Part (b): How high above the building will the ball rise?
To find the maximum height reached above the building, we use the kinematic equation for displacement:
[tex]\[ h = ut + \frac{1}{2}at^2 \][/tex]
Where:
- [tex]\( h \)[/tex] is the maximum height above the building,
- [tex]\( u \)[/tex] is the initial velocity (18 m/s),
- [tex]\( t \)[/tex] is the time to reach maximum height (1.837 seconds),
- [tex]\( a \)[/tex] is the acceleration due to gravity (-9.8 m/s²).
Substituting the values:
[tex]\[ h = 18 \times 1.837 + \frac{1}{2} \times (-9.8) \times (1.837)^2 \][/tex]
[tex]\[ h \approx 33.066 - 16.535 \][/tex]
[tex]\[ h \approx 16.531 \, \text{meters} \][/tex]
Thus, the ball will rise approximately 16.531 meters above the building.
### Part (c): Compute the velocity with which it will strike the ground and the total time it is in motion
1. Velocity with which it will strike the ground:
First, we determine the total height the ball will fall from, which is the initial height of the building plus the maximum height above the building:
[tex]\[ \text{Total height fall} = 25 \, \text{meters} + 16.531 \, \text{meters} \][/tex]
[tex]\[ \text{Total height fall} = 41.531 \, \text{meters} \][/tex]
Using the kinematic equation for velocity where the fall starts from rest:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( u \)[/tex] is the initial velocity (0 m/s from the maximum height),
- [tex]\( a \)[/tex] is the acceleration due to gravity (9.8 m/s²),
- [tex]\( s \)[/tex] is the total height fall (41.531 meters).
Substituting the values:
[tex]\[ v^2 = 0 + 2 \times 9.8 \times 41.531 \][/tex]
[tex]\[ v^2 = 812.414 \][/tex]
[tex]\[ v \approx \sqrt{812.414} \][/tex]
[tex]\[ v \approx 28.531 \, \text{m/s} \][/tex]
Therefore, the velocity with which the ball will strike the ground is approximately 28.531 m/s.
2. Total time in motion:
The total time of motion includes the time to reach the maximum height and the time to fall back to the ground. We already have the time to reach maximum height as 1.837 seconds.
To find the time to fall from the maximum height to the ground, we use the kinematic equation for displacement:
[tex]\[ s = ut + \frac{1}{2}at^2 \][/tex]
Where:
- [tex]\( s \)[/tex] is the total height fall (41.531 meters),
- [tex]\( u \)[/tex] is the initial velocity (0 m/s from rest),
- [tex]\( a \)[/tex] is the acceleration due to gravity (9.8 m/s²).
Solving for [tex]\( t \)[/tex]:
[tex]\[ 41.531 = \frac{1}{2} \times 9.8 \times t^2 \][/tex]
[tex]\[ 41.531 = 4.9 \times t^2 \][/tex]
[tex]\[ t^2 \approx \frac{41.531}{4.9} \][/tex]
[tex]\[ t^2 \approx 8.478 \][/tex]
[tex]\[ t \approx \sqrt{8.478} \][/tex]
[tex]\[ t \approx 2.911 \, \text{seconds} \][/tex]
Therefore, the total time in motion is the sum of the time to reach maximum height and the time to fall back down:
[tex]\[ \text{Total time in motion} = 1.837 + 2.911 \][/tex]
[tex]\[ \text{Total time in motion} \approx 4.748 \, \text{seconds} \][/tex]
Hence, the total time it is in motion is approximately 4.748 seconds.
