A ball is thrown straight up from a bridge at a speed of 11.0 m/s. If it takes 5.5 seconds to hit the water below, what is the velocity just before it hits the water?
Hi = bridge height
0 = water height
a = g = -9.81 m/s^2
v = Vi - g t = 11 -9.81 t
0 =Hi + Vi t + (1/2) a t^2 = Hi + 11 t - 4.9 t^2
but t = 5.5
v = 11 - 9.81(5.5) = 11 - 54 = - 43 m/s
I assume there is a part b asking for Hi, the bridge height.
To determine the velocity just before the ball hits the water, we can use the equations of motion.
Step 1: Determine the initial velocity (u) of the ball.
Given: The ball is thrown straight up from a bridge at a speed of 11.0 m/s.
Step 2: Determine the time (t) it takes for the ball to hit the water below.
Given: It takes 5.5 seconds to hit the water below.
Step 3: Determine the acceleration (a) acting on the ball.
Since the ball is in free fall, the acceleration due to gravity is approximately 9.8 m/s^2, directed downwards.
Step 4: Determine the final velocity (v) of the ball.
Using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the known values:
v = 11.0 m/s + (-9.8 m/s^2) * 5.5 s
v = 11.0 m/s - 53.9 m/s
v ≈ -42.9 m/s (rounded to one decimal place)
Therefore, the velocity just before the ball hits the water is approximately -42.9 m/s. Note that the negative sign indicates that the velocity is directed downwards.
To find the velocity just before the ball hits the water, we need to use the equation of motion for vertical displacement. This equation relates the initial velocity, final velocity, time, and acceleration.
The ball is thrown straight up, meaning its initial velocity is positive (+11.0 m/s) because it's moving upwards. When it hits the water, its final velocity is negative (-v m/s) because it's moving downwards. The acceleration in the vertical direction is the acceleration due to gravity, which is approximately -9.8 m/s^2 (taking downward as negative).
Using the equation:
vf = vi + at
where:
vf = final velocity (unknown)
vi = initial velocity (11.0 m/s)
a = acceleration (-9.8 m/s^2)
t = time taken (5.5 seconds)
We can rearrange the equation to solve for vf:
vf = vi + at
vf = 11.0 m/s + (-9.8 m/s^2)(5.5 s)
vf = 11.0 m/s - 53.9 m/s
vf = -42.9 m/s
Therefore, the velocity just before the ball hits the water is approximately -42.9 m/s. Note that the negative sign indicates that the velocity is directed downward.