A rock is thrown vertically upward with a speed of 11.0 from the roof of a building that is 70.0 above the ground. Assume free fall.
In how many seconds after being thrown does the rock strike the ground?
What is the speed of the rock just before it strikes the ground?
The smart aleck answer is, "If it's thrown straigt up then how can it land on the ground? It should land on the roof, right?"
Ignoring that, figure out how long it will rise and what it's velocity will be when it returns to the building's height, then use that for the initial velocity if it had been thrown off the building.
Once again I see that you're not supplying units so I don't know what to make of the question.
refer to the other qs
To determine the time it takes for the rock to strike the ground, we can use the equation of motion for free fall.
First, let's find the time it takes for the rock to reach its maximum height. We can use the formula:
time = (final velocity - initial velocity) / acceleration
In this case, the final velocity is 0 because the rock reaches its highest point momentarily and momentarily comes to rest. The initial velocity is 11.0 m/s (given) and the acceleration is -9.8 m/s^2 (assuming acceleration due to gravity).
So, plugging in the values:
time taken to reach maximum height = (0 - 11.0) / -9.8
Simplifying the equation, we get:
time taken to reach maximum height = 11.0 / 9.8
time taken to reach maximum height ≈ 1.12 seconds
Now, since the rock is in free fall, it will take the same amount of time to come back down to the ground. So, the total time taken for the rock to strike the ground is approximately 1.12 seconds x 2 = 2.24 seconds.
To find the speed of the rock just before it strikes the ground, we can use the equation:
final velocity = initial velocity + (acceleration x time)
In this case, the initial velocity is 11.0 m/s (given), the acceleration is -9.8 m/s^2 (assuming acceleration due to gravity), and the time is 2.24 seconds (calculated earlier).
So, plugging in the values:
final velocity = 11.0 + (-9.8 x 2.24)
Simplifying the equation, we get:
final velocity = 11.0 - 21.95
final velocity ≈ -10.95 m/s
The negative sign indicates that the rock is moving in the opposite direction (downwards) just before it strikes the ground.
Therefore, the rock strikes the ground after approximately 2.24 seconds and its speed just before striking the ground is approximately 10.95 m/s (in the downward direction).