A ball is thrown vertically upward from the edge of a bridge 26.5 m high with an initial speed of 11.0 m/s. The ball falls all the way down and strikes the water below. Determine the magnitude of the velocity of the stone just before it strike the water. (In m/s)
solve for t in
26.5 + 11.0t - 4.9t^2 = 0
Then use that value for t in
v = 11.0 - 9.8
oops: 11.0 - 9.8t
To determine the magnitude of the velocity of the ball just before it strikes the water, we need to calculate the final velocity of the ball when it reaches the water level.
Step 1: Find the time it takes for the ball to reach its maximum height:
Using the formula:
v = u + at,
where:
v = final velocity (0 m/s at maximum height, as the ball momentarily stops),
u = initial velocity (11.0 m/s),
a = acceleration (acceleration due to gravity, -9.8 m/s^2),
t = time,
we can rearrange the formula to solve for t:
t = (v - u) / a.
t = (0 - 11.0) / (-9.8) = 1.12 seconds.
Step 2: Calculate the maximum height reached by the ball:
Using the formula:
h = u * t + (1/2) * a * t^2,
where:
h = maximum height,
u = initial velocity (11.0 m/s),
t = time (1.12 seconds),
a = acceleration due to gravity (-9.8 m/s^2),
we can substitute the values and solve for h:
h = 11.0 * 1.12 + (1/2) * (-9.8) * (1.12)^2
= 6.23 meters.
Step 3: Calculate the time it takes for the ball to fall from the maximum height to the water level:
Since the ball was thrown upward and the acceleration due to gravity is pointing downwards, the total displacement from the maximum height to the water level is:
26.5 + 6.23 = 32.73 meters.
Using the formula for displacement:
s = ut + (1/2) * a * t^2,
where:
s = displacement (32.73 meters),
u = initial velocity (0 m/s at the maximum height),
t = time.
We can plug in the values and solve for t:
32.73 = 0 * t + (1/2) * (-9.8) * t^2.
Rearranging the equation:
4.9 * t^2 = 32.73.
Dividing both sides by 4.9:
t^2 = 6.68.
Taking the square root of both sides:
t = √6.68 = 2.59 seconds.
Step 4: Find the final velocity of the ball just before it strikes the water:
Using the formula:
v = u + at,
where:
v = final velocity,
u = initial velocity (0 m/s),
a = acceleration due to gravity (-9.8 m/s^2),
t = time (2.59 seconds).
We can substitute the values and find the final velocity:
v = 0 + (-9.8) * 2.59
= -25.34 m/s.
However, since we are looking for the magnitude (absolute value) of the velocity, the magnitude of the final velocity is:
|v| = |-25.34| = 25.34 m/s.
Therefore, the magnitude of the velocity of the ball just before it strikes the water is approximately 25.34 m/s.
To find the magnitude of the velocity of the stone just before it strikes the water, we can use the equation for projectile motion.
First, let's break down the problem:
1. We are given the initial vertical position (26.5 m) and the initial speed (11.0 m/s) of the ball thrown vertically upward.
2. The ball will travel upward until it reaches its highest point where its velocity becomes zero, and then it will fall back down.
Now, let's find the time it takes for the ball to reach the highest point:
The formula for vertical displacement during free fall is:
Δy = v₀t + (1/2)gt²
Where:
Δy = vertical displacement
v₀ = initial velocity
t = time
g = acceleration due to gravity (approximately 9.8 m/s²)
At the highest point, the vertical displacement (Δy) is equal to zero, and the final velocity (v) is zero.
0 = (11.0 m/s)t - (1/2)(9.8 m/s²)t²
0 = 11t - 4.9t²
To solve this quadratic equation, we can factor or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± √(b² - 4ac))/(2a)
Here, a = -4.9, b = 11.0, and c = 0. Plugging in these values:
t = (-(11.0) ± √((11.0)² - 4(-4.9)(0)))/(2(-4.9))
t = (-(11.0) ± √(121 + 0))/(2(-4.9))
t = (-(11.0) ± √121)/(2(-4.9))
t = (-11.0 ± 11)/(-9.8)
Solving for t, we have two possible solutions:
1. t = (-11.0 + 11)/(-9.8) = 0 s (Ignoring this because time cannot be negative)
2. t = (-11.0 - 11)/(-9.8) ≈ 2.24 s
The time taken to reach the highest point is approximately 2.24 seconds.
Now, let's find the final velocity of the ball just before it strikes the water:
v = v₀ + gt
where:
v = final velocity
v₀ = initial velocity
g = acceleration due to gravity (approximately 9.8 m/s²)
t = time taken
Plugging in the values:
v = 11.0 m/s + (9.8 m/s²)(2.24 s)
Calculating this:
v ≈ 11.0 m/s + 22.03 m/s
v ≈ 33.03 m/s
Therefore, the magnitude of the velocity of the stone just before it strikes the water is approximately 33.03 m/s.