The answers to the questions of the object thrown off a cliff with a horizontal speed of 10 m/s and that reach the ground which is 30 m below after 3 seconds, are:
a. We know:
b. We need to find the initial vertical velocity and the total horizontal distance traveled.
c. To calculate the initial vertical velocity and the total horizontal distance traveled by the object, we need to use the following equations:
[tex] y_{f} = y_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2} [/tex] (1)
[tex] v_{i_{x}} = \frac{x}{t} [/tex] (2)
d. The initial vertical velocity and the horizontal distance traveled by the object are 4.72 m/s and 30 m, respectively.
a. From the statement, we know:
b. We need to find the initial vertical velocity and the total horizontal distance traveled.
c. To calculate the initial vertical velocity we can use the following equation:
[tex] y_{f} = y_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2} [/tex] (1)
Where:
[tex] y_{f}[/tex]: is the final height = 0 m
[tex] y_{i}[/tex]: is the initial height = 30 m
[tex] v_{i_{y}}[/tex]: is the initial vertical velocity =?
g: is the acceleration due to gravity = 9.81 m/s²
t: is the time = 3 s
And, to find the horizontal distance traveled by the object we need to use the equation:
[tex] v_{i_{x}} = \frac{x}{t} [/tex] (2)
Where:
[tex]v_{i_{x}} [/tex]: is the initial horizontal velocity = 10 m/s
x: is the horizontal distance =?
d. The solution to this problem is the following.
[tex] y_{f} = y_{i} + v_{i_{y}}t - \frac{1}{2}gt^{2} [/tex]
[tex] 0 = 30 m + v_{i_{y}}*3 s - \frac{1}{2}*9.81 m/s^{2}*(3 s)^{2} [/tex]
Solving for [tex]v_{i_{y}}[/tex]
[tex] v_{i_{y}} = \frac{\frac{1}{2}*9.81 m/s^{2}*(3 s)^{2} - 30 m}{3 s} = 4.72 m/s [/tex]
Hence, the initial vertical velocity is 4.72 m/s.
[tex] v_{i_{x}} = \frac{x}{t} [/tex]
[tex] 10 m/s = \frac{x}{3 s} [/tex]
[tex] x = 10 m/s*3 s = 30 m [/tex]
Therefore, the horizontal distance traveled by the object is 30 m.
Read more here:
I hope it helps you!
