a stone is thrown straight upward and rises to a maximum vertical height of 40.0m. with what velocity was it thrown and how long does it take for the stone to return to the initial level from which it was thrown?
Ignore air resistance => no energy loss
PE at 40.0 m = mgh = 40*9.8m J
=392 J
PE=KE=(1/2)mv²
vi=sqrt(2gh)=sqrt(2*9.8*40)
= 28 m/s
Time = 2 * time it takes to reach highest point
= 2*(28/9.8)
= 5.71 s
To find the initial velocity and the time it takes for the stone to return to its initial level, we can use the equations of motion for vertical motion.
1. Initial velocity (u):
- When the stone reaches its maximum height, its vertical velocity becomes zero.
- At the maximum height, the stone has only the initial vertical velocity component, which is on the way up.
- We can use the equation v = u + at, where v is the final velocity (zero at the top), u is the initial velocity, a is acceleration (acceleration due to gravity, -9.8 m/s²), and t is the time taken.
- Since the stone is thrown straight upward, the final velocity at the highest point is zero.
- Rearranging the equation gives u = -at.
- Substituting a = -9.8 m/s², we have u = -(-9.8)t = 9.8t.
- The initial velocity of the stone is 9.8 times the time taken to reach the maximum height.
2. Time to return to the initial level:
- The time taken for the stone to rise to the maximum height and return to the initial level is the total time of flight.
- At the highest point, the vertical velocity component becomes zero.
- We can use the equation v = u + at, where v is the final velocity, u is the initial velocity (9.8t), a is acceleration (-9.8 m/s²), and t is the total time taken.
- Rearranging the equation gives v = u + at => 0 = 9.8t - 9.8t.
- The total time taken for the stone to return to the initial level is zero.
Therefore, the initial velocity of the stone is 9.8t, and the time taken for the stone to return to its initial level is zero.