A rock is thrown vertically upward from the ground with an initial speed 15m/s.
a. how high does go?
b. how much time is required for the rock to reach its maximum height?
c. what is the rock's height at t=2.00s?
To answer these questions, we can use the equations of motion for vertical motion under constant acceleration. Let's break down the steps to find the answers.
a. To find how high the rock goes, we need to determine its maximum height, which is the highest point it reaches in its trajectory. At the highest point, the vertical velocity becomes zero. We can use the equation:
vf = vi + at
Where:
vf = final velocity (which is 0 m/s at the highest point)
vi = initial velocity (15 m/s)
a = acceleration (acceleration due to gravity, which we assume as -9.8 m/s^2 since it acts downwards)
Rearranging the equation to solve for time (t) at the highest point:
0 = 15 - 9.8t
9.8t = 15
t = 15 / 9.8
t ≈ 1.53 seconds
Now, we know that it takes approximately 1.53 seconds for the rock to reach its maximum height.
Next, we can calculate the maximum height reached by the rock using the equation of motion:
vf^2 = vi^2 + 2as
Where:
vf = final velocity (0 m/s at the highest point)
vi = initial velocity (15 m/s)
a = acceleration (-9.8 m/s^2)
s = displacement (maximum height)
Rearranging the equation to solve for the maximum height (s):
0 = (15^2) + 2(-9.8)s
0 = 225 - 19.6s
-225 = -19.6s
s ≈ 11.47 meters
Therefore, the rock reaches a maximum height of approximately 11.47 meters.
b. We already determined earlier that it takes approximately 1.53 seconds for the rock to reach its maximum height.
c. To find the rock's height at t = 2.00 seconds, we can use the equation:
s = vi*t + (1/2)*a*t^2
Where:
s = displacement (height at time t)
vi = initial velocity (15 m/s)
t = time (2.00 s)
a = acceleration (-9.8 m/s^2)
Plugging in the values:
s = 15*2 + (1/2)*(-9.8)*(2)^2
s = 30 - 9.8*2
s = 30 - 19.6
s ≈ 10.4 meters
Therefore, at t = 2.00 seconds, the rock's height is approximately 10.4 meters.