You throw a glob of putty straight up toward the ceiling. Which is 3.60 m above the point where the putty leaves your hand. The initial speed of the putty leaves as it leaves your hand is 9.50m/s.<br>
(A) what is the speed of the putty just before it strikes the ceiling?<br>
(B) how much time from when it leaves your hand does it take the putty to reach the ceiling?
A. V^2 = Vo^2 + 2g*h = 9.5^2 - 19.6*3.6 = 19.69.
V = 4.44 m/s.
B. V = Vo + g*t = 4.44.
9.5 - 9.8t = 4.44.
-9.8t = -5.06.
t = 0.533 s.
(A) Well, since the putty is going straight up, it's eventually gonna come back down, right? So right before it strikes the ceiling, its speed would be the same as its initial speed, which is 9.50 m/s. It's like the putty is saying, "Hey, I'm coming back down, but before I do, let's party at 9.50 m/s!"
(B) Ah, the timeless question of timing! To figure out how much time it takes for the putty to reach the ceiling, we can use a little physics magic. We know the initial speed of the putty is 9.50 m/s and the distance is 3.60 m. We can use the equation: time = distance / speed. Plugging in the values, we get: time = 3.60 m / 9.50 m/s. Solving that, we find that it takes approximately 0.38 seconds for the putty to reach the ceiling. So you better be quick with your party balloons and confetti!
To solve this problem, let's use the equations of motion for vertical motion.
We are given the following information:
Initial speed, u = 9.50 m/s (upwards)
Displacement, s = 3.60 m
Acceleration due to gravity, g = 9.8 m/s² (downwards)
(A) To find the speed of the putty just before it strikes the ceiling, we can use the equation:
v² = u² + 2as
where v is the final velocity.
Substituting the given values, we have:
v² = (9.50 m/s)² + 2(-9.8 m/s²)(3.60 m)
v² = 90.25 m²/s² - 70.56 m²/s²
v² = 19.69 m²/s²
Taking the square root of both sides, we find:
v ≈ 4.43 m/s
Therefore, the speed of the putty just before it strikes the ceiling is approximately 4.43 m/s.
(B) To find the time it takes for the putty to reach the ceiling, we can use the equation:
v = u + at
where a is the acceleration due to gravity, and t is the time taken.
Rearranging the equation to solve for t, we have:
t = (v - u) / a
Substituting the given values, we have:
t = (4.43 m/s - 9.50 m/s) / -9.8 m/s²
t = (-5.07 m/s) / -9.8 m/s²
t ≈ 0.52 s
Therefore, the time it takes for the putty to reach the ceiling is approximately 0.52 seconds.
To solve this problem, we can use the equations of motion under constant acceleration. In this case, the acceleration due to gravity will act in the opposite direction to the motion of the putty. The initial velocity will be 9.50 m/s, and the final velocity will be zero since the putty comes to a stop at the ceiling.
(A) To find the speed of the putty just before it strikes the ceiling, we can use the equation:
v^2 = u^2 + 2as
Where:
v = final velocity (which is zero)
u = initial velocity (9.50 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
s = displacement (3.60 m)
Rearranging the equation, we have:
0 = (9.50)^2 + 2(-9.8)(3.60)
Simplifying the equation gives us:
0 = 90.25 - 70.56
Therefore, the speed of the putty just before it strikes the ceiling is the square root of 19.69 m/s.
(B) To find the time it takes for the putty to reach the ceiling, we can use the equation:
v = u + at
Where:
v = final velocity (which is zero)
u = initial velocity (9.50 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation, we have:
0 = 9.50 - 9.8t
Simplifying the equation gives us:
9.8t = 9.50
Therefore, the time it takes for the putty to reach the ceiling is t = 9.50 / 9.8 = 0.9694 seconds.