The boy on the tower of height h = 18 m in the figure below throws a ball a distance of x = 62 m, as shown. At what speed, in m/s, is the ball thrown?
To find the speed at which the ball is thrown, we can use the equations of motion.
The key equation to use is the horizontal motion equation:
x = v(initial) * t + (1/2) * a * t^2
In this equation, x is the horizontal distance traveled, v(initial) is the initial velocity of the ball, t is the time of flight, and a is the acceleration (which is 0 since there is no horizontal force acting on the ball).
In this case, the given horizontal distance is x = 62 m. We need to find the time of flight, t.
To find the time of flight, we can use the vertical motion equation:
h = (1/2) * g * t^2
In this equation, h is the vertical distance (height) the ball is thrown from, g is the acceleration due to gravity (approximated as 9.8 m/s^2), and t is the time of flight.
In this case, the given height is h = 18 m. We can rearrange the equation to solve for t:
t^2 = (2 * h) / g
t^2 = (2 * 18) / 9.8
t^2 = 3.67
t ≈ 1.92 s
Now that we have the time of flight, we can use the horizontal motion equation to find the initial velocity, v(initial):
x = v(initial) * t
62 = v(initial) * 1.92
v(initial) ≈ 32.29 m/s
Therefore, the ball is thrown at a speed of approximately 32.29 m/s.