The boy on the tower of height h = 20 m in the figure below throws a ball a horizontal distance of x = 56 m. At what speed, in m/s, is the ball thrown?

The product of the horizontal velocity at which it is thrown (Vx) and the time required to fall 20 m (T) equals the horizontal distance travelled, which you have been given.

Vx = 56 m/T

(1/2)gT^2 = 20
T = sqrt(40m/g)

Solve for the horizontal velocity

29

sorry don't listen to that answer

To find the speed at which the ball is thrown, we can use the equations of motion.

Firstly, let's consider the vertical motion of the ball. Since we know the height of the tower (h = 20 m), we can use the equation of motion for vertical displacement:

h = ut + (1/2)gt^2
where:
h = vertical displacement (20 m)
u = initial vertical velocity (0 m/s, since the ball is thrown horizontally)
g = acceleration due to gravity (9.8 m/s^2)
t = time of flight

We can rearrange this equation to solve for t:

20 = (1/2)(9.8)t^2
40 = 9.8t^2
t^2 = 40 / 9.8
t = √(40 / 9.8)
t ≈ 2.03 seconds

Now, let's consider the horizontal motion of the ball. The horizontal distance traveled by the ball (x = 56 m) can be found using the equation of motion for horizontal displacement:

x = ut
where:
x = horizontal displacement (56 m)
u = initial horizontal velocity

Since the ball is thrown horizontally, the initial horizontal velocity is the same as the final horizontal velocity. Therefore, we can rewrite this equation as:

x = vt

Now, we can solve for the horizontal velocity (v):

v = x / t
v = 56 m / 2.03 s
v ≈ 27.59 m/s

So, the ball is thrown at a speed of approximately 27.59 m/s.