Answer :
Given that the acceleration due to gravity [tex]\( g = -9.8 \, \text{m/s}^2 \)[/tex], we start with the equations of motion.
### Part a: Finding the velocity of the object for all relevant times.
Start with the acceleration equation:
[tex]\[ a(t) = v'(t) = g \][/tex]
Since [tex]\( g = -9.8 \, \text{m/s}^2 \)[/tex]:
[tex]\[ a(t) = v'(t) = -9.8 \][/tex]
Integrate the acceleration with respect to time to find the velocity function [tex]\( v(t) \)[/tex]:
[tex]\[ v(t) = \int -9.8 \, dt = -9.8t + C \][/tex]
Given that the initial velocity [tex]\( v_0 = 33 \, \text{m/s} \)[/tex], we can solve for the constant [tex]\( C \)[/tex]:
[tex]\[ v(0) = 33 \Rightarrow 33 = -9.8(0) + C \][/tex]
[tex]\[ C = 33 \][/tex]
Thus, the velocity function is:
[tex]\[ v(t) = 33 - 9.8t \][/tex]
### Part b: Finding the position of the object for all relevant times.
Now, integrate the velocity function to find the position function [tex]\( s(t) \)[/tex]:
[tex]\[ s(t) = \int (33 - 9.8t) \, dt = 33t - \frac{9.8t^2}{2} + D \][/tex]
Given that the initial position [tex]\( s_0 = 0 \)[/tex] (starting from the ground):
[tex]\[ s(0) = 0 \Rightarrow 0 = 33(0) - \frac{9.8 (0)^2}{2} + D \][/tex]
[tex]\[ D = 0 \][/tex]
Thus, the position function is:
[tex]\[ s(t) = 33t - 4.9t^2 \][/tex]
### Part c: Finding the time when the object reaches its highest point and the height at that time.
The highest point is reached when the velocity is zero:
[tex]\[ v(t) = 0 \Rightarrow 33 - 9.8t = 0 \][/tex]
[tex]\[ 9.8t = 33 \][/tex]
[tex]\[ t = \frac{33}{9.8} \][/tex]
[tex]\[ t \approx 3.367 \, \text{seconds} \][/tex]
To find the height at this time, substitute [tex]\( t = 3.367 \)[/tex] back into the position function:
[tex]\[ s(3.367) = 33(3.367) - 4.9(3.367)^2 \][/tex]
[tex]\[ s(3.367) \approx 55.561 \, \text{meters} \][/tex]
Hence, the highest point is reached at [tex]\( t \approx 3.367 \, \text{seconds} \)[/tex] and the height is approximately [tex]\( 55.561 \, \text{meters} \)[/tex].
### Part d: Finding the time when the object strikes the ground.
The object strikes the ground when the position is zero again:
[tex]\[ s(t) = 0 \Rightarrow 33t - 4.9t^2 = 0 \][/tex]
[tex]\[ t(33 - 4.9t) = 0 \][/tex]
This gives two solutions:
[tex]\[ t = 0 \quad (\text{initial launch time}) \][/tex]
[tex]\[ 33 - 4.9t = 0 \][/tex]
[tex]\[ t = \frac{33}{4.9} \][/tex]
[tex]\[ t \approx 6.735 \, \text{seconds} \][/tex]
So, the object strikes the ground at [tex]\( t \approx 6.735 \, \text{seconds} \)[/tex].
### Summary
a. The velocity function is:
[tex]\[ v(t) = 33 - 9.8t \][/tex]
b. The position function is:
[tex]\[ s(t) = 33t - 4.9t^2 \][/tex]
c. The object reaches its highest point at [tex]\( t \approx 3.367 \, \text{seconds} \)[/tex], and the height at this time is approximately [tex]\( 55.561 \, \text{meters} \)[/tex].
d. The object strikes the ground at [tex]\( t \approx 6.735 \, \text{seconds} \)[/tex].
### Part a: Finding the velocity of the object for all relevant times.
Start with the acceleration equation:
[tex]\[ a(t) = v'(t) = g \][/tex]
Since [tex]\( g = -9.8 \, \text{m/s}^2 \)[/tex]:
[tex]\[ a(t) = v'(t) = -9.8 \][/tex]
Integrate the acceleration with respect to time to find the velocity function [tex]\( v(t) \)[/tex]:
[tex]\[ v(t) = \int -9.8 \, dt = -9.8t + C \][/tex]
Given that the initial velocity [tex]\( v_0 = 33 \, \text{m/s} \)[/tex], we can solve for the constant [tex]\( C \)[/tex]:
[tex]\[ v(0) = 33 \Rightarrow 33 = -9.8(0) + C \][/tex]
[tex]\[ C = 33 \][/tex]
Thus, the velocity function is:
[tex]\[ v(t) = 33 - 9.8t \][/tex]
### Part b: Finding the position of the object for all relevant times.
Now, integrate the velocity function to find the position function [tex]\( s(t) \)[/tex]:
[tex]\[ s(t) = \int (33 - 9.8t) \, dt = 33t - \frac{9.8t^2}{2} + D \][/tex]
Given that the initial position [tex]\( s_0 = 0 \)[/tex] (starting from the ground):
[tex]\[ s(0) = 0 \Rightarrow 0 = 33(0) - \frac{9.8 (0)^2}{2} + D \][/tex]
[tex]\[ D = 0 \][/tex]
Thus, the position function is:
[tex]\[ s(t) = 33t - 4.9t^2 \][/tex]
### Part c: Finding the time when the object reaches its highest point and the height at that time.
The highest point is reached when the velocity is zero:
[tex]\[ v(t) = 0 \Rightarrow 33 - 9.8t = 0 \][/tex]
[tex]\[ 9.8t = 33 \][/tex]
[tex]\[ t = \frac{33}{9.8} \][/tex]
[tex]\[ t \approx 3.367 \, \text{seconds} \][/tex]
To find the height at this time, substitute [tex]\( t = 3.367 \)[/tex] back into the position function:
[tex]\[ s(3.367) = 33(3.367) - 4.9(3.367)^2 \][/tex]
[tex]\[ s(3.367) \approx 55.561 \, \text{meters} \][/tex]
Hence, the highest point is reached at [tex]\( t \approx 3.367 \, \text{seconds} \)[/tex] and the height is approximately [tex]\( 55.561 \, \text{meters} \)[/tex].
### Part d: Finding the time when the object strikes the ground.
The object strikes the ground when the position is zero again:
[tex]\[ s(t) = 0 \Rightarrow 33t - 4.9t^2 = 0 \][/tex]
[tex]\[ t(33 - 4.9t) = 0 \][/tex]
This gives two solutions:
[tex]\[ t = 0 \quad (\text{initial launch time}) \][/tex]
[tex]\[ 33 - 4.9t = 0 \][/tex]
[tex]\[ t = \frac{33}{4.9} \][/tex]
[tex]\[ t \approx 6.735 \, \text{seconds} \][/tex]
So, the object strikes the ground at [tex]\( t \approx 6.735 \, \text{seconds} \)[/tex].
### Summary
a. The velocity function is:
[tex]\[ v(t) = 33 - 9.8t \][/tex]
b. The position function is:
[tex]\[ s(t) = 33t - 4.9t^2 \][/tex]
c. The object reaches its highest point at [tex]\( t \approx 3.367 \, \text{seconds} \)[/tex], and the height at this time is approximately [tex]\( 55.561 \, \text{meters} \)[/tex].
d. The object strikes the ground at [tex]\( t \approx 6.735 \, \text{seconds} \)[/tex].
