"a ball is thrown upwards with initial velocity 20 m/s. After reaching maximum height, on the way down it strikes a bird which is 10 m above the ground. What is the velocity of the ball when it hits the bird?"
I know acceleration would be -9.8 m/s^2 (to assume free fall), the initial velocity would be 20, and that the final position would be 10, but I don't know how to find the max height to then use to find the final velocity. thank you
Just set this up as a quadratic function. I assume that we start from a height of zero, since you don't say.
h(t) = 20t - 4.9t^2
Now, you want to solve
20t - 4.9t^2 = 10
t = 0.583, 3.498
Since v=20-9.8t, that would make
v(3.498) = 20-9.8*3.498 = -14.28
You don't need to find the max height for this problem, but that occurs when v=0. So, h(2) = 20.4
Ohhh ok. Thank you so much! :D
To find the maximum height reached by the ball, we can use the kinematic equation:
v_f^2 = v_i^2 + 2aΔy
Where:
v_f is the final velocity (0 m/s at the maximum height)
v_i is the initial velocity (20 m/s)
a is the acceleration (-9.8 m/s^2)
Δy is the change in height (unknown)
Substituting the given values into the equation:
0^2 = 20^2 + 2(-9.8)Δy
0 = 400 - 19.6Δy
19.6Δy = 400
Δy = 400/19.6
Δy ≈ 20.41 m
So, the maximum height reached by the ball is approximately 20.41 meters.
Now, to find the velocity of the ball when it hits the bird, we can use the kinematic equation:
v_f = v_i + at
Where:
v_f is the final velocity (unknown)
v_i is the initial velocity (20 m/s)
a is the acceleration (-9.8 m/s^2)
t is the time it takes for the ball to travel from the maximum height to the bird (unknown)
Since the ball is thrown upwards and then falls downwards, the time taken to reach the maximum height is equal to the time taken to fall from the maximum height to the bird.
Using the equation:
Δy = v_i*t + 0.5*a*t^2
Substituting the values:
20.41 = 20*t + 0.5*(-9.8)*t^2
20.41 = 20t - 4.9t^2
Rearranging the equation:
4.9t^2 - 20t + 20.41 = 0
Solving this quadratic equation, we find two possible values for t: t = 2.065 s and t = 0.954 s. However, we discard the negative value since time cannot be negative in this context.
Using t = 2.065 s, we can now find the final velocity:
v_f = v_i + at
v_f = 20 + (-9.8) * 2.065
v_f = -0.5 m/s
Therefore, the velocity of the ball when it hits the bird is approximately -0.5 m/s.
To find the maximum height reached by the ball, you can use the kinematic equation for vertical motion:
v_f^2 = v_i^2 + 2aΔy
Where:
v_f = final velocity
v_i = initial velocity
a = acceleration
Δy = change in position
In this case, we can assume that the ball starts with an initial velocity of 20 m/s and accelerates downwards due to gravity with an acceleration of -9.8 m/s^2.
At the maximum height, the final velocity will be momentarily zero before the ball starts descending. Thus, we can rearrange the equation as follows:
0 = (20)^2 + 2(-9.8)Δy
Now, let's solve for Δy, which represents the maximum height:
0 = 400 - 19.6Δy
19.6Δy = 400
Δy = 400 / 19.6
Δy ≈ 20.41 meters
Therefore, the maximum height reached by the ball is approximately 20.41 meters. Now, we can calculate the final velocity when the ball hits the bird.
To find the final velocity, we will use the equation:
v_f = v_i + at
Substituting the known values:
v_f = 20 + (-9.8)t
Since the ball starts from rest at the maximum height, the initial velocity (v_i) is zero. We can now calculate the time it takes for the ball to fall from the maximum height to the height of the bird, which is 20.41 - 10 = 10.41 meters.
10.41 = (-9.8)t
Solving for t:
t = 10.41 / (-9.8)
t ≈ -1.06 seconds
Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball takes approximately 1.06 seconds to fall from the maximum height to hit the bird.
Now, we can calculate the final velocity (v_f):
v_f = 20 + (-9.8)(-1.06)
v_f ≈ 20 + 10.368
v_f ≈ 30.37 m/s
Hence, the velocity of the ball when it hits the bird is approximately 30.37 m/s.