1.A stone fired straight upwards from the top a 50m high bridge fall into water below it 12seconds later.what was its initial velocity? how high above the bridge did it go?
0 = 50 + Vi t - 4.9 t^2
0 = 50 + Vi (12) - 4.9 (144)
12 Vi = 705.6 - 50 = 655.6
Vi = 54.6 m/s
0 = 54.6 - 9.8 t
t = 5.57 seconds rising
h above bridge = 0 + Vi t - 4.9 t^2
= 54.6(5.57) - 4.9 (5.57)^2
= 152 m
To find the initial velocity and the maximum height above the bridge reached by the stone, we can use the equations of motion.
1. Finding the initial velocity (u):
We know that the stone is dropped from a height of 50m. Using the equation for freefall:
s = ut + (1/2)gt^2
Where:
s = displacement (50m)
u = initial velocity
t = time (12s)
g = acceleration due to gravity (typically taken as 9.8 m/s^2)
Rearranging the equation, we solve for u:
50 = u(12) + (1/2)(9.8)(12^2)
50 = 12u + 705.6
12u = -655.6
u = -54.63 m/s
The negative sign indicates that the stone was going upward (opposite to the direction we've taken as positive) when it was dropped.
So, the initial velocity of the stone was approximately -54.63 m/s.
2. Finding the maximum height above the bridge (h):
We can use the equation:
v^2 = u^2 + 2gh
Where:
v = final velocity (0 m/s, as it stops at the highest point)
h = maximum height
Rearranging the equation, we solve for h:
0 = (-54.63)^2 + 2(9.8)h
h = [(54.63)^2] / (2 * 9.8)
h ≈ 153.63 m
Therefore, the stone's maximum height above the bridge was approximately 153.63 meters.