A boy drops a stone-A, down from a cliff 480 meters high, next he throws a stone-B, 1 second later - at what initial velocity was the stone-B be thrown so as to meet the stone-A on ground at the same time? g=10 m/s2.
Require the following:
(g/2)T^2 = 480
(g/2)(T-1)^2 + V*(T-1) = 480
You have two equations in two unknowns (V and T) and can solve for V.
The first equation tells you that stone A requires
T = 9.8 seconds to fall to the ground. (if g= 10)
Use that value of T in the second equation to solve for V.
To find the initial velocity of stone-B, we can use the equations of motion for both stones and set their displacements equal to each other.
Let's break down the problem step by step:
1. For stone-A:
- Initial velocity (uA) = 0 m/s (as the stone is dropped from rest)
- Acceleration (a) = g = 10 m/s^2 (as the stone falls freely under gravity)
- Displacement (sA) = -480 m (negative since the stone is falling downwards)
2. For stone-B:
- Acceleration (a) = g = 10 m/s^2 (assuming the stone also falls freely under gravity)
- Displacement (sB) = 0 m (since stone-B is thrown horizontally, it does not have any vertical displacement)
Using the equation of motion: s = ut + (1/2)at^2
For stone-A:
- sA = 0 + (1/2) * (-10) * (tA)^2
- -480 = -5tA^2
For stone-B:
- sB = uB * tB + (1/2) * 10 * (tB)^2
- 0 = uB * tB + 5 * (tB)^2
Since both stones will meet on the ground, their time of flight (tA and tB) would be the same. Hence, we can consider tA and tB as t.
Combining both equations:
- -480 = -5t^2 [Equation 1]
- 0 = uB * t + 5t^2 [Equation 2]
To solve these equations simultaneously, we can substitute equation 1 into equation 2:
- 0 = uB * t + 5(-480 / -5)
- 0 = uB * t + 480
Simplifying:
- uB * t = -480
- uB = -480 / t [Equation 3]
Now, as we know that tA = tB, let's find the value of t.
Using the equation of motion: s = ut + (1/2)at^2 for stone-A:
- sA = 0 + (1/2) * (-10) * (tA)^2
- -480 = -5tA^2
- tA^2 = 96
Taking the square root of both sides:
- tA = √96
- tA ≈ 9.798 s
Since tA ≈ tB = 9.798 s, we can substitute this value back into Equation 3 to find the initial velocity of stone-B:
- uB = -480 / tB
- uB = -480 / 9.798
- uB ≈ -49.02 m/s
Therefore, the initial velocity at which stone-B must be thrown, in order to meet stone-A on the ground at the same time, is approximately -49.02 m/s. The negative sign indicates that stone-B is thrown in the opposite direction to the motion of stone-A.