Two Object Were Projected Vertically upward at different time with speed of 80m/s and 100m/s respectively. if the time interval is 2s. when and where will they meet .
which one starts first...
surely the slower one is shot first (per the "respectively")
set their heights equal to find the time when they meet:
80t - 4.9t^2 = 100(t-2) - 4.9(t-2)^2
solve for t to find when
evaluate h(t) to find where
extra credit: is the first object rising or falling when they meet?
if the faster one goes 1st, they don't meet in flight
huh - I would not have guessed that.
In fact, if the slower one goes second, it needs to have an initial speed of at least 90.2 m/s for it to stay aloft long enough to meet the faster one before it hits in 20.48 seconds.
To find when and where the two objects will meet, we can start by calculating the time it takes for each object to reach its maximum height.
Let's assume the object projected with a speed of 80 m/s reaches its maximum height first.
For an object moving vertically upward, the time taken to reach the maximum height is equal to the initial velocity divided by the acceleration due to gravity (g).
Given:
Initial velocity (u1) = 80 m/s
Acceleration due to gravity (g) = 9.8 m/s²
To find the time taken by the first object to reach maximum height, we can use the formula:
Time (t1) = u1 / g
Plugging in the values, we get:
t1 = 80 m/s / 9.8 m/s²
t1 = 8.16 s (approximately)
Now, let's calculate the time taken by the second object to reach its maximum height.
Given:
Initial velocity (u2) = 100 m/s
Acceleration due to gravity (g) = 9.8 m/s²
To find the time taken by the second object to reach maximum height, we use the formula:
Time (t2) = u2 / g
Plugging in the values, we get:
t2 = 100 m/s / 9.8 m/s²
t2 = 10.20 s (approximately)
Since the time interval between the two objects is given as 2 seconds, the second object is in the air for 10.20 seconds - 2 seconds = 8.20 seconds.
Now, in these 8.20 seconds, the first object has already returned to the ground and started moving downward due to gravity. So, we need to calculate where and when the second object will meet the descending first object.
Since both objects are moving vertically, and the second object is projected after the first one, the point of their meeting will be the maximum height reached by the second object.
Using simple kinematic equations, we can find the maximum height (h) reached by the second object.
Given:
Initial velocity (u2) = 100 m/s
Time (t2) = 8.20 s
Using the equation:
h = u2 * t2 - (1/2) * g * t2^2
Plugging in the values, we get:
h = 100 m/s * 8.20 s - (1/2) * 9.8 m/s² * (8.20 s)^2
h = 820 m - 329.144 m
h ≈ 490.856 m
Therefore, the two objects will meet at a height of approximately 490.856 meters from the ground.