Help imediately please very urgent!
A stone is dropped from a cliff 200m high,if a second stone is thrown vertically upwards 1.50sec after the first was released,strike the base of the cliff at the same in stant as the first stone,with what velocity was the second stone thrown?
To solve this problem, we can start by finding the time it takes for the first stone to hit the base of the cliff. We can use the formula for the time of free fall:
t = sqrt((2h) / g)
Where:
t = time of free fall
h = height of the cliff = 200m
g = acceleration due to gravity = 9.8 m/s^2 (assuming it's on Earth)
Plugging in the values:
t = sqrt((2 * 200) / 9.8) ≈ 6.42 seconds
So, it takes approximately 6.42 seconds for the first stone to hit the base of the cliff.
Now, let's find the velocity with which the second stone was thrown upwards. We can use the formula for vertical displacement:
h = v * t + (1/2) * g * t^2
Where:
h = height of the cliff = 200m
v = initial velocity of the second stone (upwards)
t = time taken for the second stone to hit the base = 6.42 seconds
g = acceleration due to gravity = 9.8 m/s^2 (assuming it's on Earth)
Plugging in the values:
200 = v * 6.42 + (1/2) * 9.8 * (6.42)^2
Now, we can rearrange the equation to solve for the initial velocity (v):
v * 6.42 = 200 - (1/2) * 9.8 * (6.42)^2
v ≈ (200 - (1/2) * 9.8 * (6.42)^2) / 6.42
By evaluating the right side of the equation, we find that the velocity with which the second stone was thrown upwards is approximately equal to 31.488 m/s.