WA ball is tossed from an upper-story window of a building. The ball is given an initial velocity of
8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far
horizontally from the base of the building does the ball strike the ground? (b) Find the height from
which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the
level of launching?
To solve this problem, we can use the kinematic equations of motion.
(a) To find how far horizontally the ball strikes the ground, we can use the equation for horizontal motion:
Horizontal distance = horizontal velocity * time
Given:
Initial horizontal velocity (horizontal component of 8 m/s at an angle of 20 degrees below the horizontal):
Vx = 8 cos(20) = 7.53 m/s
Time taken to strike the ground: t = 3.00 s
Therefore, horizontal distance = 7.53 m/s * 3.00 s = 22.59 meters
So, the ball strikes the ground 22.59 meters horizontally from the base of the building.
(b) To find the height from which the ball was thrown, we can use the equation for vertical motion:
Vertical distance = initial vertical velocity * time + 0.5 * acceleration * time^2
Given:
Acceleration due to gravity, g = 9.81 m/s^2
Initial vertical velocity (vertical component of 8 m/s at an angle of 20 degrees below the horizontal):
Vy = 8 sin(20) = 2.72 m/s
Using the above equation, we can find the height from which the ball was thrown:
Vertical distance = 2.72 m/s * 3.00 s + 0.5 * 9.81 m/s^2 * (3.00 s)^2 = 12.30 meters
So, the ball was thrown from a height of 12.30 meters.
(c) To find how long it takes the ball to reach a point 10.0 m below the level of launching, we can use the equation for vertical motion:
Vertical distance = initial vertical velocity * time + 0.5 * acceleration * time^2
Given:
Vertical distance = 10.0 m (negative since it's below the launching point)
Using the above equation and rearranging it to solve for time, we get:
-10.0 m = 2.72 m/s * t + 0.5 * 9.81 m/s^2 * t^2
0 = 4.905 t^2 + 2.72 t + 10
This is a quadratic equation that can be solved using the quadratic formula.
The solutions for t will give us the time it takes for the ball to reach a point 10.0 m below the launching point.
Solving the quadratic equation, we find:
t = (-2.72 ± sqrt((2.72)^2 - 4*4.905*10)) / (2 * 4.905)
t ≈ 1.52 s (ignoring the negative solution)
Therefore, it takes approximately 1.52 seconds for the ball to reach a point 10.0 m below the level of launching.