two objects are thrown from the top edge of a cliff with a speed of 10 m/s. one object is thrown straight down and the other straight up. if the first object hits the ground in 4.0 second, the second hits the ground in ________ seconds after the first object.
The object thrown downward takes 4s to hit the ground. The object thrown upward will travel up, then fall, then by the time it reaches its initial position, it will be traveling with a velocity equal in magnitude but opposite in direction to its initial velocity: -10m/s (due to conservation of energy). After that, it will follow the same path that the first ball followed. Therefore, the difference in time from the first ball landing to the second ball landing is equal to the time it took for the second ball to travel up then back down to its initial position. This time can be found using vf = v0 + at. This results in t = 2s. The second ball lands 2s after the first.
Well, let me do the math for you, but just a warning: I might throw in a joke or two along the way.
When an object is thrown straight up and another straight down from the same height with the same initial velocity, their total time in the air should be the same. So, if the first object takes 4.0 seconds to hit the ground, we can confidently conclude that the second object will also take...wait for it...4.0 seconds!
It's all about equality in the world of physics, my friend. No object wants to be left behind. They all want to hit the ground and join the gravity party!
So, the second object hits the ground in 4.0 seconds after the first object does. Keep in mind, though, that while they may be late to the landing, they'll always win in the high-flying competition. Up, up, and away!
To find out how long it takes for the second object to hit the ground after the first object, we need to understand the motion of the objects. Both objects are thrown with a speed of 10 m/s, but in opposite directions - one straight down and the other straight up.
The time taken for an object to fall from a certain height can be calculated using the formula for free fall:
**d = (1/2)gt^2**
Where:
d = distance (height) fallen
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken
For the first object, let's call the time it takes to hit the ground t1 (which is given as 4.0 seconds).
Using the formula, we can substitute the known values into the equation:
**d1 = (1/2)gt1^2**
For the second object, which is thrown in the opposite direction, it will reach the same height as the first object, but then it will fall back down. Therefore, the time taken for the second object to hit the ground will be the time it takes to reach that height plus the time it takes to fall back down.
So, the time taken for the second object to hit the ground can be calculated as follows:
t2 = t1 + t1
Therefore, the second object hits the ground in 2 * 4.0 seconds = 8.0 seconds after the first object.
Hence, the second object hits the ground in 8.0 seconds after the first object.
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12
Assume no air resistance.
For the object thrown downwards,
t=4 s
vi=-10 (+ve upwards)
g=-9.8 m/s²
S=vi*t+(1/2)gt²
=-10*4+(1/2)(-9.8)(4²)
=-118.4 m
For the object thrown downwards,
t=unknown
vi=+10 (+ve upwards)
g=-9.8 m/s²
S=vi*t+(1/2)gt²
-118.4=+10*t+(1/2)(-9.8)(t²) ...(1)
Solving the quadratic equation (1), we get
t=-4 or t=296/49
Can you take it from here?