A stone was thrown into the air and returned after 4.2 seconds.
a) At what speed was the stone thrown? (20.6) <- i got that right.
b) How high did the stone go? (21.6 m)
i don't know how to get b. what steps do i do?
The stone with an initial velocity, v, has a kinetic energy of (1/2)mv².
When at the top, velocity = 0, but the potential energy is mgh.
So equate energies to get
mgh = (1/2)mv²
Solve for h.
To find the height the stone went, you can use the equation of motion for an object in free fall:
h = v0t + (1/2)gt^2
where:
- h is the height
- v0 is the initial velocity
- t is the time
- g is the acceleration due to gravity (which is approximately -9.8 m/s^2 near the Earth's surface)
In this case, we know that the stone was thrown upward and then returned to its original height, so its initial velocity (v0) was positive.
Let's break down the steps to find the height:
1. Determine the time it takes for the stone to reach its maximum height. Since the stone returns after 4.2 seconds, the time it takes to reach the highest point will be half of that time: t/2 = 4.2 s / 2 = 2.1 s.
2. Calculate the initial velocity (v0) using the formula: v0 = gt. In this case, t = 2.1 s.
v0 = (9.8 m/s^2) * (2.1 s) = 20.58 m/s (approx.)
Note: We consider v0 to be positive as the stone is thrown upward.
3. Now, use the equation of motion to determine the height (h) the stone reached. Plug in the values into the equation:
h = v0t + (1/2)gt^2
= (20.58 m/s) * (2.1 s) + (1/2)(-9.8 m/s^2)(2.1 s)^2
Calculating this, you will find:
h = 43.1 m - 21.5 m = 21.6 m (approx.)
So, the stone reached a height of approximately 21.6 meters.