Jeff tosses a can of soda pop to Karen, who is standing on her 3rd floor balcony a distance of 8.5m above Jeff’s hand. Jeff gives the can an initial velocity of 16m/s, fast enough so that the can goes up past Karen, who catches the can on its way down. Calculate the velocity of the can the instant before Karen grabs the can. How long after Jeff tosses the can does Karen have to prepare to catch it?
V^2 = Vo^2 + 2g*h.
V^2 = 16^2 - 19.6*8.5 = 89.4
V = 9.46 m/s = The velocity when it reaches the balcony and when it returns to the balcony and grabbed by Karen.
V = Vo + g*Tr = 0.
16 - 9.8Tr = 0
Tr = 1.63 s. = Rise time from Jeff's hand to max ht.
V = Vo + g*Tf = 9.46.
0 + 9.8Tf = 9.46
Tf = 0.965 s. = Fall time from max ht. to balcony where Karen grabs it.
Tr+Tf = 1.63 + 0.965 = 4.23 s. = Time in air = Time Karen have to prepare to catch it.
To find the velocity of the can the instant before Karen grabs it, we need to use the equations of motion. We can use the equation for final velocity (v) in terms of initial velocity (u), acceleration (a), and distance (s):
v^2 = u^2 + 2as
Given:
u (initial velocity) = 16 m/s (upwards)
a (acceleration) = -9.8 m/s^2 (acceleration due to gravity)
s (distance) = 8.5 m
Since we want to find the velocity at the instant Karen grabs the can, the final distance is 0. Therefore, we can rewrite the equation as:
v^2 = u^2 + 2as
0 = 16^2 + 2(-9.8)(8.5)
Simplifying the equation:
0 = 256 - 166.6
166.6 = 256
v^2 = 166.6
Taking the square root of both sides:
v = √166.6
v ≈ 12.91 m/s
Therefore, the velocity of the can the instant before Karen grabs it is approximately 12.91 m/s.
To find out how long after Jeff tosses the can Karen has to prepare to catch it, we can use the equation of motion for the time (t) taken to travel a certain distance (s):
s = ut + (1/2)at^2
Given:
u (initial velocity) = 16 m/s
a (acceleration) = -9.8 m/s^2
s (distance) = 8.5 m
Again, since we want to find the time when the distance is 0 (when Karen grabs it), we can rewrite the equation as:
0 = 16t + (1/2)(-9.8)t^2
Simplifying the equation:
0 = 16t - 4.9t^2
Rearranging the terms:
4.9t^2 - 16t = 0
Factoring out t:
t(4.9t - 16) = 0
Either t = 0 or 4.9t - 16 = 0
Solving for t:
4.9t - 16 = 0
4.9t = 16
t ≈ 3.27 seconds
Therefore, Karen has approximately 3.27 seconds to prepare to catch the can.
To calculate the velocity of the can the instant before Karen grabs it, we need to analyze the motion of the can. We can break this motion into two separate parts: upward motion and downward motion.
First, let's calculate the time it takes for the can to reach its maximum height. We can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Considering the upward motion, the final velocity (v) is 0 m/s (since the can reaches its maximum height and then starts to descend). The initial velocity (u) is 16 m/s, and the acceleration (a) is the acceleration due to gravity, which is approximately -9.8 m/s² (taking downward as positive).
Plugging these values into the equation, we have:
0 = 16 + (-9.8)t.
Solving for t, we find:
t = 16/9.8 s.
The time it takes for the can to reach its maximum height is approximately 1.63 seconds.
Next, we need to calculate the time it takes for the can to fall from its maximum height to Karen's hand. Since the can is already at its maximum height, the initial velocity is 0 m/s, and the acceleration is still -9.8 m/s².
We'll use the following equation to find the time taken for the downward motion:
s = ut + (1/2)at²,
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
The distance the can needs to cover is the vertical distance between Jeff's hand and Karen's 3rd floor balcony, which is 8.5 meters.
Plugging in the values, we have:
8.5 = 0t + (1/2)(-9.8)t².
Simplifying the equation, we get:
4.9t² = 8.5.
Solving for t, we find:
t² = 8.5/4.9.
t ≈ √(8.5/4.9) ≈ 1.29 s.
The time it takes for the can to fall from its maximum height to Karen's hand is approximately 1.29 seconds.
Now that we have the times for the upward and downward motions, we can add them together to find the total time it takes for the can to reach Karen's hand:
Total time = time for upward motion + time for downward motion
Total time ≈ 1.63 s + 1.29 s
Total time ≈ 2.92 s.
Therefore, Karen has approximately 2.92 seconds to prepare to catch the can after Jeff tosses it.
To calculate the velocity of the can just before Karen grabs it, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Considering the downward motion, the initial velocity (u) is 0 m/s (since the can starts falling from rest at the top), the acceleration (a) is -9.8 m/s² (acceleration due to gravity, taking downward as positive), and the time (t) is the total time calculated earlier, which is approximately 2.92 seconds.
Plugging these values into the equation, we have:
v = 0 + (-9.8)(2.92).
Solving for v, we find:
v ≈ -28.57 m/s.
The velocity of the can just before Karen grabs it is approximately -28.57 m/s. The negative sign indicates that the velocity is directed downward, as the can is moving downward at that moment.