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

To find the answers to the given questions, we need to use the equations of motion and apply them to each stone separately. Let's break down the steps to get the answers:

(a) The time taken for the first stone to fall 75m:
We can use the equation of motion: h = ut + (1/2)gt^2, where
- h is the height (75m),
- u is the initial velocity (which is 0 since the stone is dropped),
- g is the acceleration due to gravity (approximately 9.8 m/s^2), and
- t is the time taken.

Rearranging the equation, we get: t = √(2h/g).
Substituting the given values, we have:
t = √(2 * 75 / 9.8)
t ≈ √(15.306)
t ≈ 3.91 seconds

Therefore, the time taken for the first stone to fall 75m is approximately 3.91 seconds.

(b) The time taken for the second stone to hit the ground:
Since both stones hit the ground at the same time, the time taken for the second stone to hit the ground will be the same as the time taken for the first stone. So, it is also approximately 3.91 seconds.

(c) Finding the initial velocity of the second stone:
To find the initial velocity of the second stone, we need to consider the motion of the first stone and calculate the velocity it had reached after falling 15m.

We can use the equation of motion: v = u + gt, where
- v is the final velocity (0 m/s since the stone hits the ground),
- u is the initial velocity of the first stone (which we need to find), and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values, we have:
0 = u + 9.8 * 3.91
u = -9.8 * 3.91
u ≈ -38.26 m/s

Therefore, the initial velocity of the second stone should be approximately 38.26 m/s, directed downward.

Note: The negative sign indicates that the velocity is in the opposite direction of the chosen positive direction (downward, in this case).

To solve this problem, we can use the equations of motion to determine the time taken and initial velocity of the second stone.

(a) Time taken for the first stone to fall 75m:
We can use the equation for free fall:
h = 0.5 * g * t^2
where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.

Plugging in the values:
75 = 0.5 * 9.8 * t^2
Simplifying the equation:
t^2 = 75 / (0.5 * 9.8)
t^2 = 15.31
Taking the square root of both sides:
t ≈ √15.31
t ≈ 3.92 seconds

(b) Time taken for the second stone to hit the ground:
Since both stones hit the ground at the same time, the time taken for the second stone will also be 3.92 seconds.

(c) Initial velocity of the second stone:
To find the initial velocity of the second stone, we can use the equation:
v = u + g*t
where v is the final velocity (0 m/s at the ground), u is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.

Plugging in the values:
0 = u + 9.8 * 3.92
Simplifying the equation:
u = -9.8 * 3.92
u ≈ -38.38 m/s (rounded to two decimal places)

Therefore, the initial velocity of the second stone is approximately -38.38 m/s. Note that the negative sign denotes that the stone was thrown downward.

a. d = 0.5gt^2,

75 = 0.5 * 9.8t^2,
75 = 4.9t^2,
Divide both sides by 4.9:
t^2 = 75 / 4.9 = 15.3,
t = sqrt(15.3) = 3.9 s.

b. d = Vo*t + 0.5gt^2,
Vo*t = 15 m,
75 = 15 + 0.5 * 9.8t^2,
75 - 15 = 4.9t^2,
60 = 4.9t^2,
Divide both sides by 4.9:
t^2 = 60 / 4.9 = 12.2,
t = sqrt(12.2) = 3.5 s.

c. Vo*t = 15,
Vo * 3.5 = 15,
Vo = 15 / 3.5 = 4.3 m/s.