An object is projected vertically upwards with a velocity of 28m/s. Two seconds later, another object is thrown vertically upwards with a velocity of 32 m/s. When and where will the two objects meet?
But you don't know at how many seconds the first object is projected upwards! Wouldn't you sub in those motion formulas for this type of question?
To find when and where the two objects will meet, we need to consider their respective equations of motion.
Let's start with the first object. We can use the equation of motion for vertical motion:
\[v = u + at\]
where:
v = final velocity (which is 0 when the object reaches its peak)
u = initial velocity (28 m/s, since the object was projected upwards)
a = acceleration due to gravity (-9.8 m/s^2, since the object is moving against gravity)
t = time
Rearranging the equation, we get:
\[t = \frac{(v - u)}{a}\]
Substituting the values, we can calculate that the first object will take approximately 2.86 seconds to reach its peak.
Now, let's move on to the second object. Using the same equation of motion, we have:
\[v = u + at\]
where:
v = 0 (when the object reaches its peak)
u = 32 m/s (initial velocity of the second object)
a = -9.8 m/s^2 (acceleration due to gravity)
t = time
Again, rearranging the equation, we get:
\[t = \frac{(v - u)}{a}\]
Substituting the values, we find that the second object will reach its peak in approximately 3.27 seconds.
Since the two objects meet at the highest point of their respective paths, we conclude that they will meet approximately 2.86 seconds after the second object is thrown, or 3.27 seconds after the first object was projected upwards.
To find the height at which they meet, we need to use the equation for vertical displacement:
\[s = ut + \frac{1}{2}at^2\]
For the first object, substituting the known values, we have:
\[s_1 = (28)(2.86) + \frac{1}{2}(-9.8)(2.86)^2\]
Calculating this, we find that the first object reaches a height of approximately 45.41 meters when they meet.
For the second object, substituting the known values, we have:
\[s_2 = (32)(3.27) + \frac{1}{2}(-9.8)(3.27)^2\]
Calculating this, we find that the second object reaches the same height of approximately 45.41 meters when they meet.
Therefore, the two objects will meet approximately 2.86 seconds after the second object is thrown, at a height of 45.41 meters.
Think about it.
In 2 seconds, the first object, traveling at 28m/s, is 56m high
Once the second object is thrown up at the speed of 32m/s, it will close in on the first object at the net speed of 32 - 28 = 4m/s.
You can now determine the time it takes for the second object to catch up to the first object and how far they have traveled.