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 30.2 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?
PLEASE HELP!!!!!!! :)
find the time it takes the ball to get to the bottom
Hf=hi+vi*t-1/2 g t^2
0=24.4+12*t-4.8 t^2
use the quadratic equation to find time of fall t.
Now, how fast does the person run? v=30.2/timet
To solve this problem, we need to find the time it takes for the ball to fall from the top of the building to the ground. We can use the kinematic equation for vertical displacement:
s = ut + (1/2)gt^2
where:
s = vertical displacement (24.4 m)
u = initial velocity (12 m/s, since the ball is thrown upward)
g = acceleration due to gravity (-9.8 m/s^2, assuming the positive direction is upward)
t = time
Rearranging the equation to solve for time:
t = (v - u) / g
Plugging in the values:
t = (0 - 12) / -9.8
t = 1.22 s
So, it takes approximately 1.22 seconds for the ball to fall to the ground.
Now, we need to find the average speed of the person running on the ground. We can calculate the time it takes for the person to reach the location where the ball lands.
Since the person is running at a constant speed, we can define the average speed as the total distance traveled divided by the total time taken.
The distance traveled by the person is 30.2 m.
The total time taken is the time it takes for the ball to fall, which is 1.22 s.
Average speed = total distance traveled / total time taken
Average speed = 30.2 m / 1.22 s
Average speed ≈ 24.754 m/s
Therefore, the person must have an average speed of approximately 24.754 m/s in order to catch the ball at the bottom of the building.