While standing at the edge of the roof of a building, you throw a stone upward with an initial speed of 7.65 m/s. The stone subsequently falls to the ground, which is 12.7 m below the point where the stone leaves your hand. At what speed does the stone impact the ground? How much time is the stone in the air? Ignore air resistance and take g = 9.81 m/s2. (This is NOT a suggestion to carry out such an experiment!)
V = Vo + g*Tr = 0
7.65 - 9.81*Tr = 0
9.81Tr = 7.65
Tr = 0.780 s. = Rise time.
h = ho + Vo*Tr + 0.5g*Tr^2
h = 12.7 + 7.65*0.78 - 4.9*0.78^2 = 15.69 m. Above gnd.
a. V^2 = Vo^2 + 2g*h = 0 + 19.6*15.69 = 307.4
V = 17.53 m/s
b. h = 0.5g*t^2 = 15.69
4.9t^2 = 15.69
t^2 = 3.20
t = 1.79 s. = Fall time(Tf).
Tr+Tf = 0.780 + 1.79 = 2.57 s. In air.
To find the speed at which the stone impacts the ground, and the time it takes for the stone to reach the ground, we can use the principles of kinematics. Let's break down the problem into three parts: the stone's upward motion, the stone at its maximum height, and the stone's downward motion.
1. Upward motion:
Given:
- Initial speed (u) = 7.65 m/s
- Gravity (g) = 9.81 m/s^2 (downward direction)
Since the stone is moving upwards against gravity, we can use the equation for vertical displacement during uniformly accelerated motion:
s = u*t - (1/2)*g*t^2
In this case, the initial displacement (s) is 0 because the stone starts from rest at the edge of the roof.
At maximum height, the stone momentarily stops before falling back down. Therefore, the stone reaches its maximum height when its vertical velocity (v) becomes zero.
Using the equation for final velocity in uniformly accelerated motion:
v = u - g*t
Substituting v = 0, we can solve for the time it takes for the stone to reach its maximum height.
0 = u - g*t_max_height
t_max_height = u / g
2. At maximum height:
The stone momentarily stops at the maximum height. Therefore, its vertical velocity (v) is zero. Using the equation for final velocity in uniformly accelerated motion, we can find the time it takes for the stone to fall from the maximum height to the ground.
Using the equation:
v = u - g*t
Substituting the known values:
0 = u - g*t_return_to_ground
3. Downward motion:
We can use the equation for vertical displacement again, but this time with known initial velocity and time taken to return to the ground.
s = u*t - (1/2)*g*t^2
Substituting the given values:
s = 0 - (1/2)*g*t_return_to_ground^2
Since s = -12.7 m (negative sign indicates downward direction), we can solve for t_return_to_ground.
Now we can calculate the speed at which the stone impacts the ground by using the equation for final velocity in uniformly accelerated motion:
v = u - g*t_return_to_ground
Let's plug in the values and calculate:
u = 7.65 m/s
g = 9.81 m/s^2
t_max_height = u / g
t_return_to_ground = sqrt(-2*s / g)
v = u - g*t_return_to_ground
By substituting the given values, we can find the speed at which the stone impacts the ground and the time it spends in the air.