A rock thrown straight up climbs for 2.50 s,the falls to the ground. Neglecting the air resistance, with wat velocity did the rock strike the ground?
The same speed that it was thrown, but travelling in the opposite direction. You can conclude that from an energy conservation argument.
The kinetic energy when thrown and landing equals the potential energy at the highest point of the trajectory.
(1/2) M V^2 = M g H
The time of flight T satisfies
H = (V/2)T
where T is the time of flight.
Thus V^2 = 2 g H = g T V
V = g T
There is an easier way to reach the same result, but I didn't see it.
T = V/g is the time it takes for the rock to reach zero velocity, which it does at the top of the trajectory.
Well, I hate to break it to you, but rocks aren't known for their swimming abilities, so I don't think the rock struck the ground with any velocity. It probably struck it with a loud "ouch" sound instead.
To determine the final velocity of the rock when it strikes the ground, we need to consider the motion of the rock in two phases: when it is thrown upwards and when it falls back downwards.
Phase 1: Upward motion
Since the rock is thrown straight up, it initially accelerates due to the force of gravity in the opposite direction. We can use the equation:
vf = vi + at
where
vf = final velocity of the rock (which is 0 m/s when it reaches its highest point)
vi = initial velocity of the rock (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken to reach the highest point (2.5 s)
Rearranging the equation, we have:
vi = vf - at
= 0 - (-9.8 * 2.5)
= 0 + 24.5
= 24.5 m/s
So, the initial velocity of the rock when it is thrown upwards is 24.5 m/s.
Phase 2: Downward motion
When the rock falls back downwards, it will accelerate due to gravity in the same direction. The final velocity reached during this phase will be the same magnitude but in the opposite direction compared to the velocity in phase 1. Since the acceleration due to gravity is constant, the time taken for the downward motion will also be equal to 2.5 seconds.
Therefore, the final velocity of the rock when it strikes the ground will also be -24.5 m/s (negative indicating the direction is downward).
Hence, the rock will strike the ground with a velocity of -24.5 m/s.
To determine the velocity with which the rock strikes the ground, we can break it down into two parts: the upward motion and the downward motion.
First, let's calculate the initial velocity when the rock was thrown straight up. We know that the upward motion lasts for 2.50 seconds, and during this time, the rock's velocity decreases due to the effects of gravity. At the top of its trajectory, the velocity becomes zero. We can use the equation for velocity to find the initial velocity:
v = u + at
where v is the final velocity (zero in this case), u is the initial velocity, a is the acceleration (due to gravity, -9.8 m/s^2), and t is the time (2.50 seconds).
0 = u + (-9.8 × 2.50)
Simplifying the equation, we have:
-9.8 × 2.50 = u
u = -24.5 m/s
The negative sign indicates that the initial velocity is in the opposite direction of the upward motion.
Next, we need to consider the downward motion of the rock. When it falls, it accelerates due to gravity. The time it takes to fall back to the ground is also 2.50 seconds. Since the acceleration due to gravity is constant, we can use another equation of motion:
v = u + at
where v is the final velocity (which is what we need to find), u is the initial velocity (-24.5 m/s), a is the acceleration (-9.8 m/s^2), and t is the time (2.50 seconds).
Plugging in these values, we have:
v = -24.5 + (-9.8 × 2.50)
v = -24.5 - 24.5
v = -49 m/s
Therefore, the rock strikes the ground with a velocity of 49 m/s in the downward direction.