A stone is thrown straight up from the ground with an initial speed of 30.5 m/s . At the same instant, a stone is dropped from a height of h meters above ground level. The two stones strike the ground simultaneously.Find the height h. gravity is 9.8 m/s2 .
Answer in units of m
The acceleration of
t(up) = (Vf - Vo) / g,
t(up) = (0 - 30.5) / -9.8 = 3.11m.
t(dn) = t(up) = 3.11s.
T=t(up) + t(dn) = 3.11 + 3.11 = 6.22s =
Time in flight.
The 2 stones were in flight for the same period of time:
T1 = T2 = 6.22s.
h = Vo*t + 4.9t^2,
h = 0 + 4.9*(6.22)^2 = 189.6m.
To find the height h, we can use the kinematic equation:
h = (1/2)gt^2
where g is the acceleration due to gravity (9.8 m/s^2) and t is the time taken for the stone to reach the ground.
The time taken for the stone to reach the ground can be determined using the equation:
t = (2v)/g
where v is the initial velocity of the stone.
Given that the stone is thrown straight up with an initial speed of 30.5 m/s, we can substitute this value into the equation:
t = (2 * 30.5 m/s) / 9.8 m/s^2
t ≈ 6.22 s
Now, we can substitute this value of t into the equation for h:
h = (1/2) * 9.8 m/s^2 * (6.22 s)^2
h ≈ 191.18 m
Therefore, the height h is approximately 191.18 meters.
To solve this problem, we can use the equations of motion for both the stone thrown upwards and the stone dropped.
For the stone thrown upwards, we have:
Initial velocity (u) = 30.5 m/s (upwards)
Initial position (s1) = 0 m (since it starts from the ground level)
Acceleration (a) = -9.8 m/s^2 (negative sign as it is acting in the opposite direction to the initial velocity)
Time (t) = unknown
Final position (s2) = unknown (but we know it will be the height h above the ground level when it strikes the ground)
Using the equation: s2 = s1 + u*t + 0.5*a*t^2 and substituting the given values, we get:
h = 0 + (30.5)t + 0.5*(-9.8)t^2
For the stone dropped, we have:
Initial velocity (u) = 0 m/s (as it is dropped from rest)
Initial position (s1) = h m (since it is dropped from a height of h above ground level)
Acceleration (a) = 9.8 m/s^2 (positive sign as it is acting in the same direction as the initial position)
Time (t) = unknown
Final position (s2) = 0 m (since it strikes the ground)
Using the equation: s2 = s1 + u*t + 0.5*a*t^2 and substituting the given values, we get:
0 = h + 0 + 0.5*(9.8)*t^2
Now, since the stones strike the ground simultaneously, the time taken for both stones should be the same. Hence, we can equate the two equations:
h = 0 + (30.5)t + 0.5*(-9.8)t^2 = h + 0 + 0.5*(9.8)*t^2
Simplifying the equation, we get:
30.5t - 4.9t^2 = 0
Factorizing the equation, we get:
t(30.5 - 4.9t) = 0
This equation has two solutions:
t = 0 (which corresponds to the initial instant when the stones are released)
30.5 - 4.9t = 0 (which is the solution we are interested in)
Solving the second equation for t, we get:
4.9t = 30.5
t = 6.22 seconds (rounded to two decimal places)
Substituting t back into either of the original equations, we can find the height h:
h = 0 + (30.5)(6.22) + 0.5*(-9.8)(6.22)^2
Calculating this, we find:
h ≈ 573.79 meters
Therefore, the height h is approximately 573.79 meters.