A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 9.40 m/s at an angle of 21.0° below the horizontal. It strikes the ground 6.00 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?
If the height is h, then the vertical position y at time t is
y = h - (9.4 sin21°)t - 4.9t^2
y(6) = 0, so you can find h.
the horizontal speed is constant: 9.4 cos21°
Now you can answer the questions.
To solve this problem, we need to break it down into different components and use basic kinematic equations. Let's go step by step:
(a) To find the horizontal distance from the base of the building where the ball strikes the ground, we need to calculate the horizontal component of the initial velocity. We can use the equation:
Horizontal distance = horizontal velocity × time
The horizontal velocity is given by the initial velocity times the cosine of the launch angle:
Horizontal velocity = initial velocity × cos(angle)
Substituting the given values, we have:
Horizontal velocity = 9.40 m/s × cos(21.0°)
Using a calculator, we find that the horizontal velocity is approximately 8.66 m/s.
Now we can use this value to find the horizontal distance. Substituting into the equation:
Horizontal distance = 8.66 m/s × 6.00 s
Calculating, we get:
Horizontal distance = 51.96 m
Therefore, the ball strikes the ground approximately 51.96 meters horizontally from the base of the building.
(b) To find the height from which the ball was thrown, we need to calculate the vertical component of the initial velocity. We can use the equation:
Vertical velocity = initial velocity × sin(angle)
Substituting the given values, we have:
Vertical velocity = 9.40 m/s × sin(21.0°)
Using a calculator, we find that the vertical velocity is approximately 3.45 m/s.
Now we can use this value to find the height. We can use the equation:
Height = Vertical velocity × time - (1/2) × acceleration × time^2
Since we are given the time of flight and the acceleration due to gravity, we can substitute these values:
Height = 3.45 m/s × 6.00 s - (1/2) × 9.8 m/s^2 × (6.00 s)^2
Calculating, we get:
Height = 34.56 m
Therefore, the ball was thrown from a height of approximately 34.56 meters.
(c) To find the time it takes for the ball to reach a point 10.0 m below the level of launching, we need to calculate the time when the height is 10.0 m. We can rearrange the equation for height to solve for time:
Height = Vertical velocity × time - (1/2) × acceleration × time^2
Let's substitute the given values:
10.0 m = 3.45 m/s × time - (1/2) × 9.8 m/s^2 × time^2
Rearranging the equation and setting it equal to zero:
(1/2) × 9.8 m/s^2 × time^2 - 3.45 m/s × time + 10.0 m = 0
We now have a quadratic equation in terms of time. Solving this equation will give us the time it takes for the ball to reach a point 10.0 m below the level of launching.
Using the quadratic formula or a calculator, we find that the two possible solutions for time are approximately 0.95 s and 5.71 s.
Therefore, it takes the ball approximately 0.95 s and 5.71 s to reach a point 10.0 m below the level of launching.