A ball is thrown upward from the top of a 24.4 m tall building. The ball's initial speed is 12 m/s. At the same instant, a person is running on the ground at a distance of 31.4 m from the building. What must be the average speed of the person if he is to catch the ball at the bottom of the building?
Tr = (Vf-Vo)/g = (0-12) / -9.8 = 1.22 s=
Rise time or time to reach max .ht.
hmax = ho + (Vf^2-Vo^2)/2g.
hmax = 24.4 + (0-144) / -19.6 = 31.75 m=
ht. above gnd.
h = Vo*t + 0.5g*t^2 = 31.75 m.
0 + 4.9t^2 = 31.75
t^2 = 6.48
Tf = 2.55 s. = Time to fall to gnd.
D = r*(Tr+Tf).
r = D/(Tr+Tf)
r=31.4 / (1.22+2.55) = 8.33 m/s.
To find the average speed of the person, we need to determine the time it takes for the ball to reach the bottom of the building when thrown upward and then calculate the average speed.
First, let's find the time it takes for the ball to reach the bottom of the building. We can use the equation for the vertical motion of the ball:
y = v₀t - 1/2gt²
where:
- y is the vertical displacement (negative since the ball is dropped downward).
- v₀ is the initial velocity of the ball.
- t is the time taken.
- g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, the initial velocity of the ball is 12 m/s, and the vertical displacement is the height of the building, which is -24.4 m. Let's calculate the time t:
-24.4 = 12t - 1/2 * 9.8 * t²
-24.4 = 12t - 4.9t²
4.9t² - 12t - 24.4 = 0
We can solve this quadratic equation to find t. Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
where:
a = 4.9
b = -12
c = -24.4
t = (-(-12) ± √((-12)² - 4 * 4.9 * (-24.4))) / 2 * 4.9
t = (12 ± √(144 + 478.4)) / 9.8
t = (12 ± √622.4) / 9.8
Since we're looking for the time it takes for the ball to reach the bottom, we'll use the positive value in this case:
t = (12 + √622.4) / 9.8
t ≈ 2.15 s
Now that we have the time it takes for the ball to reach the bottom, we can calculate the average speed of the person. The distance the person needs to cover is 31.4 m, and the time to cover that distance is 2.15 s:
Average speed = total distance / total time
Average speed = 31.4 m / 2.15 s
Average speed ≈ 14.6 m/s
Therefore, the person must have an average speed of approximately 14.6 m/s to catch the ball at the bottom of the building.
To find the average speed of the person, we need to determine the time it takes for the ball to fall from the top of the building and calculate the average speed required to cover the horizontal distance of 31.4 m in that time.
First, let's find the time it takes for the ball to fall from the top of the building. We can use the equation of motion for the vertical motion of the ball:
h = ut + (1/2)gt^2
Where:
h = height of the building (24.4 m)
u = initial speed of the ball (12 m/s)
g = acceleration due to gravity (-9.8 m/s^2, taking into account the direction)
Rearranging the equation, we have:
t^2 - (2u/g)t - (2h/g) = 0
Solving this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 1
b = -2u/g
c = -2h/g
Substituting the values into the equation, we get:
t = (-(-2u/g) ± √((-2u/g)^2 - 4(1)(-2h/g))) / (2(1))
Simplifying further, we have:
t = (2u/g ± √(4u^2/g^2 + 8h/g)) / 2
Now we can substitute the given values:
u = 12 m/s
h = 24.4 m
g = -9.8 m/s^2
t = (2(12)/(-9.8) ± √(4(12)^2/(-9.8)^2 + 8(24.4)/(-9.8))) / 2
Calculating further:
t ≈ (24/(-9.8) ± √(576/96.04 + 195.2/(-9.8))) / 2
t ≈ (-2.45 ± √(5.99 + 19.92)) / 2
t ≈ (-2.45 ± √25.91) / 2
As time cannot be negative in this context, we can ignore the negative value.
Therefore, t ≈ (-2.45 + √25.91) / 2 ≈ (-2.45 + 5.09) / 2 ≈ 1.82 s
Now we can find the average speed required for the person to cover the horizontal distance of 31.4 m in 1.82 seconds:
Average speed = Horizontal distance / Time
Average speed = 31.4 m / 1.82 s
Calculating further:
Average speed ≈ 17.25 m/s
Therefore, the person must have an average speed of approximately 17.25 m/s to catch the ball at the bottom of the building.