a stone is dropped from a cliff 200m high.if a second stone is thrown verticall upwards 1.50seconds after the first was released.strike the cliff base at the same instant as the first stone.with what velocity was the second stone thrown
The First Stone.
d1 = 0.5g*t^2 = 4.9*(1.5)^2 = 11 m. =
Distance traveled by 1st stone after
1,5 s.
V^2 = Vo^2 + 2g*d.
V^2 = 0 + 19.6*11 = 215.6
V = 14.68 m/s = Velocity of 1st stone after 1.5 s.
d1 = Vo*t + 0.5g*t^2 = 200-11.
16.68t + 4.9t^2 = 189
4.9t^2 + 16.68t - 189 = 0.
Use Quadratic Formula.
Tf = 4.74 s. = Fall time or time for each stone to reach Gnd.
The 2nd Stone.
d = Vo*t + 0.5g*t^2 = 200 m.
Vo*4.74 + 4.9*(4.74)^2 = 200
4.74Vo + 110.1 = 200
4.74Vo = 200-110.1 = 89.9
Vo = 19 m/s.
Correction(TYPO).
Change 16.68 to 14.68. Then Tf will be
4.89 s.
Change all 4.74s to 4.89.
Vo = 89.9/4.89 = 18.4 m/s. = Initial velocity of 2nd stone.
To find the velocity with which the second stone was thrown, we can use the equations of motion.
Let's break down the problem:
1. The height from which the first stone is dropped is given as 200m.
2. We need to find the velocity with which the second stone was thrown.
3. The second stone is thrown vertically upwards 1.50 seconds after the first stone is released.
4. Both stones hit the base of the cliff at the same instant.
Now, let's use the equation of motion for the first stone:
h = ut + (1/2)gt^2
Where:
- h is the height (in this case, 200m)
- u is the initial velocity of the first stone (which is 0 as it's dropped)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time (not given)
Using this equation, we can find the time taken for the first stone to hit the ground.
200 = 0 + (1/2)(9.8)t^2
Simplifying, we get:
200 = 4.9t^2
t^2 = 200/4.9
t ≈ √(40.816)
t ≈ 6.39 s
Now, let's work with the second stone:
The second stone is thrown vertically upwards 1.50 seconds after the first stone was released. Both stones hit the base of the cliff at the same instant. This means the time taken for the second stone to hit the ground is the total time, which is 6.39 seconds.
Now, we can use the equation of motion for the second stone:
h = ut + (1/2)gt^2
Where:
- h is the height (in this case, 200 + 200m, as the stone needs to reach the top of the cliff and then fall back down)
- u is the initial velocity of the second stone (which is what we need to find)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time (6.39 seconds)
Using this equation, we can solve for the initial velocity of the second stone:
400 = u(6.39) - (1/2)(9.8)(6.39)^2
Simplifying, we get:
400 = 6.39u - 203.071
6.39u = 400 + 203.071
6.39u = 603.071
u ≈ 94.42 m/s
Therefore, the second stone was thrown with an approximate velocity of 94.42 m/s upwards.