Arann launches a rock into the air such that it rises 20.0 m in the first third (time wise) of its ascent to maximum height.
Determine the rock's maximum height and launch speed.
Determine the time(s) at which the ball is at a height of 10m.
PLEASE HELP
find the first third time
20=1/2 g t^2 + vi*t
or 4.9t^2-vi*t+20=0
and
vf=vi+gt
at the top 0=vi-9.8*3t
ok, then t=vi/(9.8*3)
Put that in the first equation, solve for vi.
Then, you can solve all else.
^^ bobpursley
What formula did you use? Could you repost it using v1 and v2 as variables?
vi is launch speed.
vf is the vertical velocity at the top,zero.
To determine the rock's maximum height and launch speed, we need to use the kinematic equations of motion.
Let's assume the initial velocity of the rock is 'u' m/s (launch speed), the maximum height it reaches is 'h' m, and the total time it takes to reach the maximum height is 't' seconds.
1. Using the first-third time, we can find the time it takes to reach the maximum height (t).
- In the first third of the ascent, the rock travels a vertical distance of 20.0 m.
- The total distance traveled during the ascent is two times the distance in the first third, which is 40.0 m (20.0 m × 2).
- The vertical motion equation is given by: S = ut + (1/2)at², where S is the vertical distance, u is initial velocity, t is time, and a is acceleration (which is -9.8 m/s² in this case, as the object is in free fall).
- Plugging in the given values, we get: -40.0 m = u(t) + (1/2)(-9.8 m/s²)t².
2. Using the equation from step 1, we can solve for time 't':
- Rearrange the equation: -4.9t² + ut + 40 = 0.
- Solve this quadratic equation for 't' using the quadratic formula:
t = [-u ± √(u² - 4(-4.9)(40))] / (2)(-4.9).
We have two potential solutions for 't': t1 and t2 (since it's a quadratic equation).
Notice that one of the values of 't' will be negative because it represents time before the rock was launched. We can discard that negative solution.
3. Now that we have the time at which the rock reaches its maximum height, we can use it to find 'h'.
- Use the equation of motion: S = ut + (1/2)at², where 'S' is the vertical distance, 'u' is initial velocity, 't' is time, and 'a' is acceleration (which is -9.8 m/s² in this case, as the object is in free fall).
- Substitute the known values: 'S = h' and 't = t1' (or t2).
- Solve the equation for 'h'.
4. To find the launch speed 'u', we can use the equation v = u + at, where 'v' is the final velocity (which is 0 at the maximum height), 'u' is initial velocity, 'a' is acceleration, and 't' is the time taken to reach the maximum height.
- Substitute the known values: 'v = 0', 'a = -9.8 m/s²', and 't = t1' (or t2).
- Rearrange the equation and solve for 'u'.
5. To determine the time(s) at which the rock is at a height of 10 m, we need to set up the equation using the equation of motion: S = ut + (1/2)at².
- We set the equation: 10 = ut + (1/2)(-9.8)t², where 'S' is the vertical distance, 'u' is the initial velocity, 't' is time, and 'a' is the acceleration.
- Solve this quadratic equation for 't' to find the time(s) at which the rock is at a height of 10 m.
Note: The calculations might involve multiple steps and could result in decimal values.