A ball is thrown horizontally from the top of a building 30.1 m high. The ball strikes the ground at a point 66.8 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s
thats not very helpful jim
To find the time the ball is in motion, we can use the equations of motion. In this case, we know the initial vertical position (y0 = 30.1 m), the final vertical position (y = 0 m), the horizontal distance (x = 66.8 m), and the acceleration due to gravity (g = 9.8 m/s^2).
Let's first find the time it takes for the ball to reach the ground vertically. We can use the formula:
y = y0 + v0y * t - (1/2) * g * t^2
Since the ball is thrown horizontally, the initial vertical velocity (v0y) is 0. Therefore, the equation simplifies to:
0 = 30.1 - (1/2) * 9.8 * t^2
Rearranging the equation, we get:
4.9 * t^2 = 30.1
Dividing both sides by 4.9, we have:
t^2 = 6.14
Next, taking the square root of both sides, we find:
t ≈ √6.14 ≈ 2.48 s
So, it takes approximately 2.48 seconds for the ball to reach the ground vertically.
Since the ball is thrown horizontally, the horizontal motion and vertical motion are independent of each other. Therefore, the time taken for the horizontal motion is the same as that for the vertical motion.
Hence, the time the ball is in motion is approximately 2.48 seconds.