A ball is thrown horizontally from the top of a building 17.6 m high. The ball strikes the ground at a point 67.7 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
To find the time the ball is in motion, we can use the kinematic equation that relates the distance, initial velocity, time, and acceleration.
In this case, the ball is thrown horizontally, so its initial vertical velocity is 0 m/s. The only force acting on the ball is gravity, causing it to accelerate downward at a rate of 9.8 m/s^2.
We can use the equation for vertical motion:
Δy = v0y * t + (1/2) * a * t^2
Where:
Δy is the vertical distance (17.6 m),
v0y is the initial vertical velocity (0 m/s),
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time in motion (what we're trying to find).
Since the initial velocity in the vertical direction is 0 m/s and the acceleration is negative due to gravity, we can rewrite the equation as:
Δy = (1/2) * a * t^2
Plugging in the given values:
17.6 m = (1/2) * (-9.8 m/s^2) * t^2
Simplifying:
17.6 m = -4.9 m/s^2 * t^2
To solve for t^2, we divide both sides of the equation by -4.9 m/s^2:
t^2 = -17.6 m / (-4.9 m/s^2)
t^2 = 3.5918
To find t, we take the square root of both sides:
t = √3.5918
Calculating the square root:
t ≈ 1.895 s
Therefore, the time the ball is in motion is approximately 1.895 seconds.
To find the time the ball is in motion, we can use the horizontal motion of the ball because gravity doesn't affect the horizontal motion.
The horizontal distance traveled by the ball is given as 67.7 m.
We know that the horizontal velocity of the ball is constant because there is no horizontal acceleration.
Therefore, we can use the equation:
distance = velocity x time
Since the ball is thrown horizontally, the initial vertical velocity is zero.
The vertical distance traveled by the ball is 17.6 m, and the acceleration in the vertical direction is due to gravity, which is -9.8 m/s^2.
Using the equation of motion:
distance = initial_velocity x time + (1/2) x acceleration x time^2
We can substitute the known values:
17.6 m = 0 x t + (1/2) x (-9.8 m/s^2) x t^2
Rearranging the equation:
17.6 m = (-4.9 m/s^2) x t^2
Dividing both sides of the equation by -4.9 m/s^2:
t^2 = -17.6 m / -4.9 m/s^2
t^2 = 3.5918 s^2
Taking the square root of both sides of the equation:
t = sqrt(3.5918 s^2)
t ≈ 1.895 s
Therefore, the time the ball is in motion is approximately 1.895 seconds.