A stone is dropped from a 75 m high building. when this stone has dropped 15m, a second stone is thrown downward with an initial velocity such that two stones hit the ground at the same time. find a. time taken for the first stone to fall 75m. (b) the time taken for the second stone to hit the ground (c.) what as the initial velocity of the second stone
hf=hi+vi*t-9.8t^2/2
both hf is the same (zero) and hi is 75.
one vi is zero.
Can you do it from here?
To solve this problem, we can use the laws of motion and equations of motion. Here's how we can find the answers to each part of the question:
(a) To find the time taken for the first stone to fall 75m, we can use the equation of motion for free fall:
\[h = \frac{1}{2} g t^2\]
where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. We can rearrange this equation to find t:
\[t = \sqrt{\frac{2h}{g}}\]
Substituting h = 75m and g ≈ 9.8 m/s^2:
\[t = \sqrt{\frac{2 \times 75}{9.8}}\]
Simplifying this equation will give us the time taken for the first stone to fall 75m.
(b) Since the second stone is thrown downward, its initial velocity will have a negative value. We can find the time taken for the second stone to hit the ground using the same equation of motion:
\[h = \frac{1}{2} g t^2\]
where h = (75 - 15) = 60m (as the second stone is thrown 15m below the 75m height of the first stone).
Substituting h = 60m and g ≈ 9.8 m/s^2, solving for t will give us the time taken for the second stone to hit the ground.
(c) To find the initial velocity of the second stone, we can use the equation of motion:
\[h = ut + \frac{1}{2} a t^2\]
where u is the initial velocity, t is the time, and a is the acceleration (which is g and negative in this case because the stone is thrown downward).
Since we have the time from part (b), we can substitute h = 60m, a ≈ -9.8 m/s^2, and solve the equation for u to find the initial velocity of the second stone.
By following these steps, we can find the answers to each part of the question.