Respuesta :
Answer:
Second stone hist at 4.08 s later.
Explanation:
The second stone straight up, so we need to find the rise and descent time.
[tex]v_{f}=v_{i}-gt[/tex]
At the top, the final speed is 0 and t is the rise time.
[tex]0=20-gt_{r}[/tex]
[tex]t_{r}=\frac{20}{9.81}[/tex]
[tex]t_{r}=2.04 s[/tex]
Let's recall that the rise time is equal to the descent time.
So the total time of the stone will be
[tex]t_{t}=4.08 s[/tex]
Now, the second stone hits the street in t2 = 4.08 s + t1 s
But we just need the time after the first stone hist the street. So we just subtract the time t1 to the second time t2, which means t2 = 4.08 s.
Therefore stone hist at 4.08 s later.
The second stone hits the wall 4.08 seconds later.
To solve this question, we will use the equation of motion. In particular, the first equation of motion which states that;
v = u + at, where
v = the final velocity
u = the initial velocity
a = acceleration of gravity
t = time taken by the stone.
From the question, we're told that the stone is thrown with a speed of 20 m/s, which is the initial velocity. And a final velocity, v of 0 m/s. Using a constant of 9.81 for acceleration of the stone. We have,
0 = 20 - 9.81 × t
20 = 9.81t
t = [tex]\frac{20}{9.81}[/tex]
t = 2.039or 2.04 seconds.
Assuming that the time taken to ascent = time taken to descent, this means that the total time taken will be 2.039 + 3.039 = 4.078 seconds or 4,08 seconds.
visit https://brainly.com/question/24040430 to learn more about the first equation
