A ball is thrown horizontally from the top of
a building 32.3 m high. The ball strikes the
ground at a point 110 m from the base of the
building.
The acceleration of gravity is 9.8 m/s
2
.
Find the time the ball is in motion.
Answer in units of s
H = 0.5g*t^2 = 32.3 m.
4.9t^2 = 32.3
t^2 = 6.59
t = 2.57 s.
To find the time the ball is in motion, we can use the equations of motion and the given information.
First, we can find the initial vertical velocity of the ball. Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s.
Next, let's find the time it takes for the ball to fall from the top of the building to the ground. We can use the equation of motion for vertical motion:
h = (1/2) * g * t^2
where h is the initial height (32.3 m), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
32.3 = (1/2) * (-9.8) * t^2
Multiplying both sides by 2:
64.6 = -9.8 * t^2
Dividing both sides by -9.8:
t^2 = 6.59693878
Taking the square root of both sides:
t = 2.566 s (rounded to 3 decimal places)
Therefore, the ball is in motion for approximately 2.566 seconds.
To find the time the ball is in motion, we can use the formula for the time of flight of a projectile. The time of flight, or the total time the ball is in motion, can be calculated using the following formula:
time = sqrt(2 * height / g)
Where:
- height is the vertical distance the ball traveled (32.3 m in this case)
- g is the acceleration due to gravity (9.8 m/s^2)
Let's substitute the values into the formula and calculate the time:
time = sqrt(2 * 32.3 / 9.8)
time = sqrt(65.69 / 9.8)
time = sqrt(6.6949)
time ≈ 2.59 seconds
Therefore, the time the ball is in motion is approximately 2.59 seconds.