a stone is from the top of a tower 300m high and at the same time another stone is projected vertically upward with initial velocity 75m/s. calculate when and where the two stone meet
d1 + d2 = 300m,
0.5gt^2 + (75t + 0.5gt^2) = 300m,
Since g in the 1st term is positive and
g in the 3rd is negative, they cancel
each other and the Eq becomes:
75t = 300,
t = 4 seconds.
d1 = 0.5*9.8*4^2 = 78.4m. below top
of tower = 300 - 78.4 = 222m above
ground.
d2 = 75*4 - 0.5*9.8*4^2,
d2 = 300 - 78.4 = 222m above ground.
Therefore, the 2 stones meet 4s after
release at a distance of 222m above
ground.
To find out when and where the two stones will meet, we need to use the equations of motion for each stone and solve the system of equations.
For the stone projected vertically upward, we can use the equation of motion:
\[ h_u = u \cdot t - \frac{1}{2}gt^2 \]
where:
- h_u is the height of the stone projected upward
- u is the initial velocity (75 m/s)
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
For the stone dropped from the top of the tower, we can use the equation of motion:
\[ h_d = \frac{1}{2}gt^2 \]
where:
- h_d is the height of the stone dropped from the tower
- g is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
Since we want to find when and where the two stones meet, we can set h_u equal to h_d:
\[ u \cdot t - \frac{1}{2}gt^2 = \frac{1}{2}gt^2 \]
Adding \(\frac{1}{2}gt^2\) to both sides:
\[ u \cdot t = gt^2 \]
Simplifying:
\[ u = gt \]
Solving for t:
\[ t = \frac{u}{g} \]
Plugging in the values:
\[ t = \frac{75\, \text{m/s}}{-9.8\, \text{m/s}^2} \]
Calculating this value gives:
\[ t \approx -7.65\, \text{s} \]
Since time cannot be negative, we discard this negative solution.
The two stones meet at time t = 7.65 s.
To find the height where they meet, we can use either \(h_u\) or \(h_d\) with this time value. Let's use \(h_u\):
\[ h_u = u \cdot t - \frac{1}{2}gt^2 \]
Plugging in the values:
\[ h_u = 75\, \text{m/s} \cdot 7.65\, \text{s} - \frac{1}{2}(-9.8\, \text{m/s}^2) \cdot (7.65\, \text{s})^2 \]
Calculating this value gives:
\[ h_u \approx 288.6\, \text{m} \]
Therefore, the two stones meet approximately 288.6 meters above the ground.