a rock is thrown vertically down from a bridge with an initial velocity of 30m/s and strikes the water in 2.0sec. find the velocity with which the rock strikes and the height of the bridge.
vi=-30m/s
a=-g=-9.8 m/s²
vf=vi+at
=-49.6 m/s
Height
=average velocity*time
=(-30-49.6)/2 * 2 sec
=-79.6 m
The negative sign means that the rock is lower than the initial position.
To solve this problem, we can utilize the equations of motion. Let's break down the given information step by step:
1. Initial velocity: The rock is thrown vertically downwards, so the initial velocity (u) is -30 m/s. The negative sign indicates the downward direction.
2. Time taken: The rock strikes the water in 2.0 seconds, so the time (t) is 2.0 s.
3. Acceleration: Since the only force acting on the rock is gravity, we can assume a constant acceleration due to gravity (g) of approximately 9.8 m/s².
Now, let's determine the velocity with which the rock strikes the water:
1. Use the equation of motion: v = u + gt, where v is the final velocity.
v = -30 m/s + (9.8 m/s²)(2.0 s)
v = -30 m/s + 19.6 m/s
v = -10.4 m/s
The negative sign indicates that the velocity is directed downward.
Thus, the velocity with which the rock strikes the water is approximately -10.4 m/s.
To find the height of the bridge, we need to use another equation of motion for vertical motion:
2. The equation for height (h) is s = ut + (1/2)gt², where s is the displacement.
Since the rock is thrown from the bridge, it starts from rest. Therefore, u = 0 m/s.
s = (1/2)gt²
s = (1/2)(9.8 m/s²)(2.0 s)²
s = (1/2)(9.8 m/s²)(4.0 s²)
s = (1/2)(9.8 m/s²)(16.0 s²)
s = 78.4 m
The height of the bridge is approximately 78.4 meters.
So, the rock strikes the water with a velocity of approximately -10.4 m/s, and the height of the bridge is approximately 78.4 meters.