A ball is thrown horizontally off the top of a 65 m tall building at a velocity of 7.8 m/s. How far did the ball land from the base of the building? Round your answer to 2 decimal places.
65 = .5 g t^2
130 = 9.8 t^2
the distance is 7.8*t
To solve this problem, we can use the kinematic equation that relates distance, initial velocity, time, and acceleration:
d = v * t
First, we need to find the time it takes for the ball to fall to the ground. Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. We can use the equation:
d = v * t + (1/2) * a * t^2
where:
- d is the vertical distance (65 meters),
- v is the initial vertical velocity (0 m/s),
- a is the acceleration due to gravity (-9.8 m/s^2),
- t is the time.
Since the ball is falling vertically, the equation becomes:
65 = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
This equation simplifies to:
-4.9 t^2 = 65
Dividing both sides by -4.9:
t^2 ≈ -13.27
Taking the square root of both sides, we get:
t ≈ √(-13.27)
Since time cannot be negative, the ball will hit the ground in approximately 3.63 seconds.
Now, let's find out how far the ball travels horizontally in that time. Since the velocity is constant throughout the motion, we can use the equation:
d = v * t
Substituting the given values:
d = 7.8 m/s * 3.63 s
Calculating this:
d ≈ 28.31 meters
Therefore, the ball will land approximately 28.31 meters from the base of the building.