Answer:
a) 3.36 sec
b) 23.9 m/s
c) 1.52 sec
Explanation:
maximum height
H = 25 +v_y^2/2g
H=25+9^2/2*9.81 =29.13 m
time taken to reach maximum height = t_1 =v_y/g = 9/9.81 =0.92 seconds
time taken to fall back to ground from maximum height = √(2gH)/g
= √(2*9.81*29.13)/9.8 = 2.439 seconds
a) Total time taken to reach ground T = t_1+t_2 =0.92+2.439=3.36 sec
b) objects final speed = √2gH = 23.9 m/s
c) let total time be t then
25 = 9t +0.5gt^2
Solving we get t= 1.52 seconds
a) The object will take a time of 3.355 seconds to reach the ground.
b) The final velocity of the object as it impacts the ground is -23.902 meters per second.
c) The object will take a time of 1.520 seconds to reach the ground.
a) The height of the object ([tex]h[/tex]), in meters, as a function of time ([tex]t[/tex]), in seconds, is described below:
[tex]y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex] (1)
Where:
Now we proceed to determine the time needed for the object to reach the ground: ([tex]y_{o} = 25\,m[/tex], [tex]y = 0\,m[/tex], [tex]v_{o} = 9\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]-4.904\cdot t^{2}+9\cdot t +25 = 0[/tex] (2)
Then, we solve this second order polynomial by quadratic formula:
[tex]t_{1} \approx 3.355\,s[/tex], [tex]t_{2} \approx -1.520\,s[/tex]
Only the former solution offers a realistic indicator. Hence, we conclude that the object will take a time of 3.355 seconds to reach the ground.
b) The final velocity of the object ([tex]v[/tex]), in meters per second, is determined by this expression:
[tex]v = v_{o}+g\cdot t[/tex] (3)
Now we proceed to find the final velocity of the object as it impact the ground: ([tex]v_{o} = 9\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]t \approx 3.355\,s[/tex])
[tex]v = 9+(-9.807)\cdot (3.355)[/tex]
[tex]v = -23.902\,\frac{m}{s}[/tex]
The final velocity of the object as it impacts the ground is -23.902 meters per second. [tex]\blacksquare[/tex]
c) In this part we shall apply the same approach in a), that is: ([tex]y_{o} = 25\,m[/tex], [tex]y = 0\,m[/tex], [tex]v_{o} = -9\,\frac{m}{s}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex])
[tex]-4.904\cdot t^{2}-9\cdot t +25 = 0[/tex]
Then, we solve this second order polynomial by quadratic formula:
[tex]t_{1} \approx 1.520\,s[/tex], [tex]t_{2} \approx -3.355\,s[/tex]
Only the former solution offers a realistic indicator. Hence, we conclude that the object will take a time of 1.520 seconds to reach the ground.
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