A stone is thrown vertically upwards with a speed of 24m/s,two seconds later a second stone is dropped from the same point.Find where the two stones meet
A ball is dropped from a height of 5m onto a concrete floor and rebounds with a speed of 0.8 times the downward speed on arrival.Find the height reached from the rebound
Let t be the time after the first stone is thrown, Y1 be the height of the first stone and Y2 the height of the dropped stone. Require that Y1 = Y2 and solve for t.
Y1 = 24t - 4.9 t^2
Y2 = -4.9(t-2)^2 = -4.9 t^2 +19.6t -19.6 (provided t > 2 s)
24t = 19.6t -19.6
There is no solution with t > 2 s
The stones never meet while in free flight
To find where the two stones meet, we need to determine the height reached by the first stone and the time it takes for the second stone to reach that height. Let's break down the problem step by step:
1. Calculate the time it takes for the first stone to reach its highest point:
- The initial velocity of the first stone is 24 m/s.
- The acceleration due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of motion).
- Using the equation: v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time), we can rearrange it as t = (v - u) / a.
- Substituting the values, t = (0 - 24) / -9.8, which gives t = 2.45 seconds.
2. Determine the maximum height reached by the first stone:
- The formula to calculate the height of an object thrown vertically upward is given by: h = ut + (1/2)at^2, where h is the height, u is the initial velocity, t is the time, and a is the acceleration.
- Substituting the values, h = 24(2.45) + (1/2)(-9.8)(2.45)^2, which gives h = 29.43 meters.
3. Find the time it takes for the second stone to fall:
- The second stone is dropped from the same height, so its initial velocity is 0 m/s.
- The acceleration due to gravity remains -9.8 m/s^2.
- We need to find the time it takes for the second stone to travel the same height as the first stone.
- Since we know the height is 29.43 meters, we can use the formula h = (1/2)at^2 and solve for t.
- Rearranging the formula, we get t = √(2h / a).
- Substituting the values, t = √(2 * 29.43 / -9.8), which gives t ≈ 2.19 seconds.
Therefore, the two stones meet approximately 2.19 seconds after the second stone is dropped from the same point.