Answer :
Let's solve each part step-by-step.
### a. Determine the velocity [tex]\( v \)[/tex] of the stone after [tex]\( t \)[/tex] seconds.
The height function given is:
[tex]\[ s(t) = -4.9t^2 + 19.6t + 24.5 \][/tex]
To find the velocity, we need to take the derivative of the height function [tex]\( s(t) \)[/tex] with respect to [tex]\( t \)[/tex]. The derivative of [tex]\( s(t) \)[/tex] is:
[tex]\[ v(t) = \frac{d}{dt} s(t) = \frac{d}{dt} \left( -4.9t^2 + 19.6t + 24.5 \right) = -9.8t + 19.6 \][/tex]
So, the object's velocity after [tex]\( t \)[/tex] seconds is
[tex]\[ v(t) = -9.8t + 19.6 \][/tex]
### b. When does the stone reach its highest point?
The stone reaches its highest point when the velocity [tex]\( v(t) \)[/tex] is zero. Setting [tex]\( v(t) = 0 \)[/tex], we get:
[tex]\[ 0 = -9.8t + 19.6 \][/tex]
Solving for [tex]\( t \)[/tex], we find:
[tex]\[ t = \frac{19.6}{9.8} = 2 \text{ seconds} \][/tex]
### c. What is the height of the stone at the highest point?
To determine the height at the highest point, we substitute [tex]\( t = 2 \)[/tex] seconds into the height function [tex]\( s(t) \)[/tex]:
[tex]\[ s(2) = -4.9(2)^2 + 19.6(2) + 24.5 \][/tex]
[tex]\[ s(2) = -4.9(4) + 39.2 + 24.5 \][/tex]
[tex]\[ s(2) = -19.6 + 39.2 + 24.5 \][/tex]
[tex]\[ s(2) = 44.1 \text{ meters} \][/tex]
### d. When does the stone strike the ground?
The stone strikes the ground when the height [tex]\( s(t) \)[/tex] is zero. We solve the equation
[tex]\[ 0 = -4.9t^2 + 19.6t + 24.5 \][/tex]
By solving this quadratic equation, we find that the time [tex]\( t \)[/tex] when the stone strikes the ground is:
[tex]\[ t = 5 \text{ seconds} \][/tex]
### e. With what velocity does the stone strike the ground?
To find the velocity when the stone strikes the ground, we substitute [tex]\( t = 5 \)[/tex] seconds into the velocity function [tex]\( v(t) \)[/tex]:
[tex]\[ v(5) = -9.8(5) + 19.6 \][/tex]
[tex]\[ v(5) = -49 + 19.6 \][/tex]
[tex]\[ v(5) = -29.4 \text{ meters per second} \][/tex]
### f. On what intervals is the speed increasing?
Speed increases when the velocity becomes more negative after reaching the highest point. The highest point occurs at [tex]\( t = 2 \)[/tex] seconds, and the stone strikes the ground at [tex]\( t = 5 \)[/tex] seconds. Therefore, the interval during which the speed is increasing is
[tex]\[ (2, 5) \][/tex]
So the answers are as follows:
a. The object's velocity after [tex]\( t \)[/tex] seconds is [tex]\( v(t) = -9.8t + 19.6 \)[/tex]
b. The stone reaches its highest point at [tex]\( t = 2 \)[/tex] seconds.
c. The height of the stone at the highest point is [tex]\( 44.1 \)[/tex] meters.
d. The stone strikes the ground at [tex]\( t = 5 \)[/tex] seconds.
e. The stone strikes the ground with a velocity of [tex]\( -29.4 \)[/tex] meters per second.
f. The speed is increasing on the interval [tex]\( (2, 5) \)[/tex].
### a. Determine the velocity [tex]\( v \)[/tex] of the stone after [tex]\( t \)[/tex] seconds.
The height function given is:
[tex]\[ s(t) = -4.9t^2 + 19.6t + 24.5 \][/tex]
To find the velocity, we need to take the derivative of the height function [tex]\( s(t) \)[/tex] with respect to [tex]\( t \)[/tex]. The derivative of [tex]\( s(t) \)[/tex] is:
[tex]\[ v(t) = \frac{d}{dt} s(t) = \frac{d}{dt} \left( -4.9t^2 + 19.6t + 24.5 \right) = -9.8t + 19.6 \][/tex]
So, the object's velocity after [tex]\( t \)[/tex] seconds is
[tex]\[ v(t) = -9.8t + 19.6 \][/tex]
### b. When does the stone reach its highest point?
The stone reaches its highest point when the velocity [tex]\( v(t) \)[/tex] is zero. Setting [tex]\( v(t) = 0 \)[/tex], we get:
[tex]\[ 0 = -9.8t + 19.6 \][/tex]
Solving for [tex]\( t \)[/tex], we find:
[tex]\[ t = \frac{19.6}{9.8} = 2 \text{ seconds} \][/tex]
### c. What is the height of the stone at the highest point?
To determine the height at the highest point, we substitute [tex]\( t = 2 \)[/tex] seconds into the height function [tex]\( s(t) \)[/tex]:
[tex]\[ s(2) = -4.9(2)^2 + 19.6(2) + 24.5 \][/tex]
[tex]\[ s(2) = -4.9(4) + 39.2 + 24.5 \][/tex]
[tex]\[ s(2) = -19.6 + 39.2 + 24.5 \][/tex]
[tex]\[ s(2) = 44.1 \text{ meters} \][/tex]
### d. When does the stone strike the ground?
The stone strikes the ground when the height [tex]\( s(t) \)[/tex] is zero. We solve the equation
[tex]\[ 0 = -4.9t^2 + 19.6t + 24.5 \][/tex]
By solving this quadratic equation, we find that the time [tex]\( t \)[/tex] when the stone strikes the ground is:
[tex]\[ t = 5 \text{ seconds} \][/tex]
### e. With what velocity does the stone strike the ground?
To find the velocity when the stone strikes the ground, we substitute [tex]\( t = 5 \)[/tex] seconds into the velocity function [tex]\( v(t) \)[/tex]:
[tex]\[ v(5) = -9.8(5) + 19.6 \][/tex]
[tex]\[ v(5) = -49 + 19.6 \][/tex]
[tex]\[ v(5) = -29.4 \text{ meters per second} \][/tex]
### f. On what intervals is the speed increasing?
Speed increases when the velocity becomes more negative after reaching the highest point. The highest point occurs at [tex]\( t = 2 \)[/tex] seconds, and the stone strikes the ground at [tex]\( t = 5 \)[/tex] seconds. Therefore, the interval during which the speed is increasing is
[tex]\[ (2, 5) \][/tex]
So the answers are as follows:
a. The object's velocity after [tex]\( t \)[/tex] seconds is [tex]\( v(t) = -9.8t + 19.6 \)[/tex]
b. The stone reaches its highest point at [tex]\( t = 2 \)[/tex] seconds.
c. The height of the stone at the highest point is [tex]\( 44.1 \)[/tex] meters.
d. The stone strikes the ground at [tex]\( t = 5 \)[/tex] seconds.
e. The stone strikes the ground with a velocity of [tex]\( -29.4 \)[/tex] meters per second.
f. The speed is increasing on the interval [tex]\( (2, 5) \)[/tex].
