A baseball is thrown horizontally from a height of 11.20 m above the ground with a speed of 27.5 m/s. Where is the ball after 0.90 s have elapsed?
The ball is 1 a horizontal distance of 2 m from the launch point.
Xo = 27.5 m/s.
Yo = 0.
T = 0.9 s. = Time in flight.
Dx = Xo*T = 27.5m/s * 0.9s = 24.75 m.
from launch point.
h = ho _ 0.5g*T^2,
h=11.2 - 4.9*(0.9)^2 =11.2-3.97=7.2 m.
above gnd.
To determine the horizontal position of the ball after 0.90 s, we can use the formula:
horizontal distance = initial horizontal velocity × time
Given that the ball is thrown horizontally, its initial horizontal velocity is equal to the speed at which it was thrown, which is 27.5 m/s.
So, the horizontal distance traveled by the ball after 0.90 s can be calculated as:
horizontal distance = 27.5 m/s × 0.90 s = 24.75 m
Therefore, the ball is located 24.75 meters horizontally from the launch point after 0.90 seconds have elapsed.
Now, let's address the second part of the question which states that the ball is 2 meters horizontally from the launch point. This information can be used to determine the time it takes for the ball to travel that distance.
Using the same formula as before, we rearrange it to solve for time:
time = horizontal distance / initial horizontal velocity
Substituting the given values, we get:
time = 2 m / 27.5 m/s ≈ 0.073 s
Therefore, it takes approximately 0.073 seconds for the ball to travel a horizontal distance of 2 meters from the launch point.