Answer:
C 1.4 s
Explanation:
Even though we have a velocity on the horizontal axis it does not affect the time to reach the ground, so we can solve this as a free-fall motion:
[tex]y=y_o-\frac{1}{2}*g*t^2\\t=\sqrt{\frac{2*10m}{9.8m/s^2}}\\t=1.4s[/tex]
so the correct answer is C, 1.4 seconds.
Answer:
The time it will take for the block to reach the ground is most nearly C. 1.4 s
Explanation:
We have the following data in order to solve the exercise :
The mass ''m'' of the block
[tex]m=2kg[/tex]
The high of the platform above a certain level which is [tex]10m[/tex]
We also know that the block is launched horizontally from the edge of the platform with an initial speed of [tex]3\frac{m}{s}[/tex]
This speed is exclusively horizontal. The vertical component of the velocity is [tex]0\frac{m}{s}[/tex]
Finally, the air resistance is negligible.
We can modeled this exercise as a free fall motion.
We only need this equation to solve the exercise :
[tex]Y_{f}(t)=Y{i}+v{i}.t+\frac{1}{2}.a.t^{2}[/tex] (I)
Where [tex]Y_{f}[/tex] is the final vertical position
Where [tex]Y_{i}[/tex] is the initial vertical position
Where [tex]v_{i}[/tex] is the initial speed
Where [tex]a[/tex] is the acceleration due to gravity which is [tex]9.81\frac{m}{s^{2}}[/tex]
And where [tex]t[/tex] is the time variable
The equation (I) modelates the vertical position in function of the variable ''t'' time.
If we use the equation (I) with some simplifications such as [tex]Y_{i}=0m[/tex] and [tex]v_{i}=0\frac{m}{s}[/tex] because the initial vertical speed is zero, we find that
[tex]10m=\frac{9.81\frac{m}{s^{2}}.t^{2}}{2}[/tex]
[tex]t^{2}=\frac{20m}{9.81\frac{m}{s^{2}}}[/tex]
[tex]t^{2}=2.0387s^{2}[/tex]
[tex]t=1.428s[/tex]
That is the time in which the block will travel [tex]10m[/tex] from the platform to the ground. The correct option is C. 1.4 s
